diff --git a/sql/task1.sql b/sql/task1.sql
index 90de336ca..4871931c6 100644
--- a/sql/task1.sql
+++ b/sql/task1.sql
@@ -1,14 +1,51 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+select *
+from `Products` p, `Categories` c
+where
+    p.category_id = c.category_id
+    and c.category_name = 'Sports & Outdoors';
+
 -- 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 user_id, username, count(order_id) as total_orders from order_data */
+select
+    order_data.user_id,
+    username,
+    count(*) as total_orders
+    from `Orders` order_data
+    LEFT JOIN `Users` user_data ON order_data.user_id = user_data.user_id
+    group by order_data.user_id, 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.
 
--- 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
+    review_data.product_id,
+    product_name,
+    avg(rating) as average_rating
+from `Reviews` review_data
+LEFT JOIN `Products` product_data ON review_data.product_id = product_data.product_id
+group by product_id, 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
+    u.user_id,
+    u.username,
+    SUM(o.total_amount) AS total_amount_spent
+FROM
+    `Orders` o 
+left JOIN
+    `Users` u ON u.user_id = o.user_id
+GROUP BY
+    u.user_id, u.username
+ORDER BY
+    total_amount_spent DESC
+LIMIT 5;
diff --git a/sql/task2.sql b/sql/task2.sql
index ad2596731..0f9430071 100644
--- a/sql/task2.sql
+++ b/sql/task2.sql
@@ -3,17 +3,76 @@
 -- 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_data.product_id,
+    product_name,
+    AVG(rating) AS average_rating
+FROM
+    `Products` product_data
+    LEFT JOIN `Reviews` review_data ON product_data.product_id = review_data.product_id
+GROUP BY
+    product_id,
+    product_name
+ORDER BY average_rating DESC
+LIMIT 1;
+
 -- 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 u.user_id, username
+FROM `Users` u
+WHERE
+    u.user_id IN (
+        SELECT O.user_id
+        FROM
+            `Orders` O
+            JOIN `Order_Items` item ON O.order_id = item.order_id
+            JOIN `Products` P ON item.product_id = P.product_id
+        GROUP BY
+            O.user_id, P.category_id
+        HAVING
+            COUNT(DISTINCT P.category_id) = (
+                SELECT COUNT(*)
+                FROM `Categories`
+            )
+    );
+
 -- 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, product_name
+FROM `Products` review_data
+WHERE
+    product_id NOT IN(
+        SELECT review_data.product_id
+        FROM review_data
+    );
+
 -- 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 u.user_id, u.username
+FROM `Users` u
+WHERE
+    u.user_id IN (
+        SELECT o.user_id
+        FROM (
+                SELECT
+                    user_id, order_date, LAG(order_date, 1) OVER (
+                        PARTITION BY
+                            user_id
+                        ORDER BY order_date
+                    ) AS prev_order_date
+                FROM `Orders` order_data
+            ) AS o
+        WHERE
+            DATEDIFF(
+                o.order_date, o.prev_order_date
+            ) = 1
+    );
\ No newline at end of file
diff --git a/sql/task3.sql b/sql/task3.sql
index f078a9439..f1152c0f0 100644
--- a/sql/task3.sql
+++ b/sql/task3.sql
@@ -2,18 +2,75 @@
 -- Write an SQL query to retrieve the top 3 categories with the highest total sales amount.
 -- 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 c.category_id, category_name, SUM(total_amount) as total_amount
+From
+    `Categories` c
+    Inner Join `Products` p ON c.category_id = p.category_id
+    Inner Join `Order_Items` oi ON oi.product_id = p.product_id
+    Inner Join `Orders` o ON oi.order_id = o.order_id
+GROUP BY
+    C.category_id
+ORDER BY total_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 u.user_id, username
+FROM
+    `Users` u
+    JOIN `Orders` o ON u.user_id = o.user_id
+    JOIN `Order_Items` oi ON o.order_id = oi.order_id
+    JOIN `Products` p ON oi.product_id = p.product_id
+    JOIN (
+        SELECT category_id
+        FROM `Categories`
+        WHERE
+            category_name = 'Toys & Games'
+    ) Toys ON p.category_id = Toys.category_id
+GROUP BY
+    u.user_id,
+    u.username,
+    Toys.category_id
+HAVING
+    COUNT(DISTINCT P.product_id) = (
+        SELECT COUNT(*)
+        FROM `Products`
+        WHERE
+            category_id = Toys.category_id
+    );
+
 -- 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 p.product_id, p.product_name, p.category_id, p.price
+FROM `Products` p
+    JOIN (
+        SELECT category_id, MAX(price) AS max_price
+        FROM `Products`
+        GROUP BY
+            category_id
+    ) P_MAX ON p.category_id = P_MAX.category_id
+    AND p.price = P_MAX.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.
+
+SELECT u.user_id, u.username
+FROM
+    `Users` u
+    JOIN `Orders` o1 on u.user_id = o1.user_id
+    JOIN `Orders` o2 on u.user_id = o2.user_id
+    AND o2.order_date = DATE_ADD(o1.order_date, INTERVAL 1 DAY)
+    JOIN `Orders` o3 on u.user_id = o3.user_id
+    AND o3.order_date = DATE_ADD(o1.order_date, INTERVAL 2 DAY)
+GROUP BY
+    u.user_id,
+    u.username
+HAVING
+    COUNT(DISTINCT o1.order_id) > 2;
\ No newline at end of file
diff --git a/tests/test_sql_queries.py b/tests/test_sql_queries.py
index 22b25d546..71bd7b716 100644
--- a/tests/test_sql_queries.py
+++ b/tests/test_sql_queries.py
@@ -1,16 +1,16 @@
 import unittest
-import psycopg2  # Replace with appropriate database connector based on your database
+import mysql.connector  # Replace with appropriate database connector based on your database
 
 class TestSQLQueries(unittest.TestCase):
 
     def setUp(self):
         # Establish a connection to your test database
-        self.conn = psycopg2.connect(
-            dbname='your_dbname',
-            user='your_username',
-            password='your_password',
-            host='your_host',
-            port='your_port'
+        self.conn = mysql.connector.connect(
+            database='shopify',
+            user='root',
+            password='1999821pjcq.',
+            host='localhost',
+            port='3306'
         )
         self.cur = self.conn.cursor()
 
@@ -21,7 +21,7 @@ def tearDown(self):
 
     def test_task1(self):
         # Task 1: Example SQL query in task1.sql
-        with open('/sql/task1.sql', 'r') as file:
+        with open('./sql/task1.sql', 'r') as file:
             sql_query = file.read()
 
         self.cur.execute(sql_query)
@@ -36,7 +36,7 @@ def test_task1(self):
 
     def test_task2(self):
         # Task 2: Example SQL query in task2.sql
-        with open('/sql/task2.sql', 'r') as file:
+        with open('./sql/task2.sql', 'r') as file:
             sql_query = file.read()
 
         self.cur.execute(sql_query)