diff --git a/sql/task1.sql b/sql/task1.sql
index 90de336ca..35406a57a 100644
--- a/sql/task1.sql
+++ b/sql/task1.sql
@@ -1,14 +1,45 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+-- Select all from products table where category id is 8 (sports).
+SELECT * FROM product_data WHERE category_id = '8';
+
 -- Problem 2: Retrieve the total number of orders for each user
 -- Write an SQL query to retrieve the total number of orders for each user.
 -- The result should include the user ID, username, and the total number of orders.
 
+-- Select columns user_id and username and count order_id from order_data
+SELECT user.user_id, user.username,
+COUNT(ord.order_id) AS total_amount FROM user_data user
+-- JOIN user_data with order_data
+JOIN order_data ord ON user.user_id = ord.user_id
+-- Group results by user ID and username.
+GROUP BY user.user_id, user.username;
+
 -- Problem 3: Retrieve the average rating for each product
 -- Write an SQL query to retrieve the average rating for each product.
 -- The result should include the product ID, product name, and the average rating.
 
+-- Select columns of product_id and product_name, and the average rating
+SELECT pr.product_id, pr.product_name,
+-- Calculate average rating
+AVG(re.rating) AS average_rating FROM product_data pr
+-- Join review_data from product_data and match product_id in both tables
+JOIN review_data re ON pr.product_id = re.product_id
+-- group by product id and product name.
+GROUP BY pr.product_id, pr.product_name;
+
 -- Problem 4: Retrieve the top 5 users with the highest total amount spent on orders
 -- Write an SQL query to retrieve the top 5 users with the highest total amount spent on orders.
 -- The result should include the user ID, username, and the total amount spent.
+
+-- Select columns of user_id and username, and sum of total_amount.
+SELECT us.user_id, us.username,
+-- Calculate sum
+SUM(ord.total_amount) AS total_amount_spent FROM user_data us
+-- join order_data with user_data and match user_id with both tables.
+JOIN order_data ord ON us.user_id = ord.user_id
+-- group results by user id and username.
+GROUP BY us.user_id, us.username
+-- order results by descending order and limit to top 5.
+ORDER BY total_amount_spent DESC LIMIT 5;
diff --git a/sql/task2.sql b/sql/task2.sql
index ad2596731..8ed28f328 100644
--- a/sql/task2.sql
+++ b/sql/task2.sql
@@ -3,17 +3,68 @@
 -- The result should include the product ID, product name, and the average rating.
 -- Hint: You may need to use subqueries or common table expressions (CTEs) to solve this problem.
 
+-- Select product id, name, and average rating.
+SELECT pr.product_id, pr.product_name, 
+-- Calculate average rating
+AVG(re.rating) AS average_rating
+-- Join review_data with product_data and match product_id in both tables.
+FROM product_data pr JOIN review_data re ON pr.product_id = re.product_id
+-- Group results by product id and name.
+GROUP BY pr.product_id, pr.product_name
+-- Filter results to show only products with the highest average rating
+HAVING AVG(re.rating) = (
+-- Create a subquery to find the highest average rating among each product.
+	SELECT MAX(average_rating) FROM (
+    -- Calculate average rating for each product.
+    SELECT AVG(rating) AS average_rating FROM review_data GROUP BY product_id
+    )
+    AS SubQuery
+    );
+
 -- Problem 6: Retrieve the users who have made at least one order in each category
 -- Write an SQL query to retrieve the users who have made at least one order in each category.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries or joins to solve this problem.
 
+-- Select user ID and username.
+SELECT us.user_id, us.username FROM user_data us
+-- Join order_data with user_data.
+JOIN order_data ord ON us.user_id = ord.user_id
+-- Join order_items_data with order_data.
+JOIN order_items_data oid ON ord.order_id = oid.order_id
+-- Join prudct_data with order_items_data.
+JOIN product_data pr ON oid.product_id = pr.product_id
+-- Group results by user_id and username.
+GROUP BY us.user_id, us.username
+-- Filter results to only show user who have ordered at least once in each category.
+HAVING COUNT(DISTINCT pr.category_id) = (
+-- Create a subquery to count the total number of distinct categories
+    SELECT COUNT(*) FROM category_data
+);
+
 -- Problem 7: Retrieve the products that have not received any reviews
 -- Write an SQL query to retrieve the products that have not received any reviews.
 -- The result should include the product ID and product name.
 -- Hint: You may need to use subqueries or left joins to solve this problem.
 
+-- Select product_id and product_name
+SELECT pr.product_id, pr.product_name FROM product_data pr
+-- Do a left join with review_data on product_id
+LEFT JOIN review_data re ON pr.product_id = re.product_id
+-- Filter results to show only products without any reviews.
+WHERE re.product_id IS NULL;
+
 -- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
 -- Write an SQL query to retrieve the users who have made consecutive orders on consecutive days.
 -- The result should include the user ID and username.
--- Hint: You may need to use subqueries or window functions to solve this problem.
\ No newline at end of file
+-- Hint: You may need to use subqueries or window functions to solve this problem.
+
+-- Select distinct user_ids and usernames.
+SELECT DISTINCT us.user_id, us.username FROM user_data us
+-- Join order_data with user_data
+JOIN order_data or1 ON us.user_id = or1.user_id
+-- Join order_data with user_data again.
+JOIN order_data or2 ON us.user_id = or2.user_id
+-- Filter results to only show order dates that are consecutive.
+WHERE DATE(or1.order_date) = date_add(date(or2.order_date), interval 1 day)
+OR DATE(or2.order_date) = date_add(date(or1.order_date), interval 1 day);
diff --git a/sql/task3.sql b/sql/task3.sql
index f078a9439..acd631427 100644
--- a/sql/task3.sql
+++ b/sql/task3.sql
@@ -3,17 +3,78 @@
 -- The result should include the category ID, category name, and the total sales amount.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+-- Select category ID, category name, and total sales amount for every category
+SELECT ca.category_id, ca.category_name, 
+-- Calculate the sum or the total sales amount
+sum(od.quantity * od.unit_price) AS total_sales_amount
+-- Join products with categories and order_items_data
+from categories ca JOIN products pr ON ca.category_id = pr.category_id
+JOIN order_items_data od ON pr.product_id = od.product_id
+-- Group results by category id and name
+GROUP BY ca.category_id, ca.category_name
+-- Order results by total sales amount in descending order and limited to the top 3 categories.
+ORDER BY total_sales_amount DESC LIMIT 3;
+
 -- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
 -- Write an SQL query to retrieve the users who have placed orders for all products in the Toys & Games
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+-- Select user ID and username
+SELECT us.user_id, us.username FROM user_data us
+-- Join user_data with order
+JOIN orders ord ON us.user_id = ord.user_id
+-- Join order_items_data with orders
+JOIN order_items_data oid ON ord.order_id = oid.order_id
+-- Join order_items_data with products
+JOIN products pr ON oid.product_id = pr.product_id
+-- Join products with categories.
+JOIN categories ca ON pr.category_id = ca.category_id
+-- Filter by Toys & Games category
+WHERE ca.category_name = 'Toys & Games' 
+-- Group results by user id and username
+GROUP BY us.user_id, us.username
+-- Filter results to show only users who have placed orders for distinct products in Toys & Games.
+HAVING COUNT(distinct pr.product_id) = (
+-- Create a subquery to count total number of distinct products in Toys & Games
+select COUNT(distinct product_id) FROM products WHERE category_id = (
+-- Create a subquery to find category_id for Toys & Games category
+SELECT category_id FROM categories WHERE category_name = 'Toys & Games'
+)
+);
+
 -- Problem 11: Retrieve the products that have the highest price within each category
 -- Write an SQL query to retrieve the products that have the highest price within each category.
 -- The result should include the product ID, product name, category ID, and price.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
 
+-- Select product id, name, category id, and price.
+SELECT pr.product_id, pr.product_name, pr.category_id, pr.price FROM products pr
+-- Join products with a subquery which finds the max price for every category. 
+INNER JOIN ( SELECT category_id, MAX(price) AS max_price FROM products 
+GROUP BY category_id )
+-- Match category id
+AS max_prices ON pr.category_id = max_prices.category_id 
+-- Match product's price with max price.
+AND pr.price = max_prices.max_price;
+
 -- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
 -- Write an SQL query to retrieve the users who have placed orders on consecutive days for at least 3 days.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
+
+-- Create a CTE that creates a dataset with user id, username, order date, previous order date and next order date.
+WITH DatesOrder AS ( SELECT ord.user_id, us.username, ord.order_date,
+-- Use lag to get previous order date for every user.
+LAG(ord.order_date, 1) OVER (partition by ord.user_id ORDER BY ord.order_date) AS
+previous_order_date,
+-- Lead provides the next order date for every user.
+LEAD(ord.order_date, 1) OVER (partition by ord.user_id ORDER BY ord.order_date) AS 
+next_order_date
+-- Join orders with user_data
+FROM orders ord JOIN user_data us ON ord.user_id = us.user_id )
+-- Select distinct user ids and usernames
+SELECT DISTINCT user_id, username
+-- Filter results to show users who have ordered on 3 consecutive days
+FROM DatesOrder WHERE order_date = date_add(previous_order_date, interval 1 DAY) AND
+next_order_date = date_add(order_date, interval 1 day);