diff --git a/cross joins lab.sql b/cross joins lab.sql
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+++ b/cross joins lab.sql	
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+-- 1 Get all pairs of actors that worked together.
+-- 2 Get all pairs of customers that have rented the same film more than 3 times.
+-- 3 Get all possible pairs of actors and films.
+
+use sakila;
+
+-- 1
+SELECT 
+    CONCAT(a1.first_name, ' ', a1.last_name) AS 'ACTOR 1',
+    CONCAT(a2.first_name, ' ', a2.last_name) AS 'ACTOR 2'
+FROM 
+    sakila.film_actor fa1
+JOIN 
+    sakila.film_actor fa2 ON fa1.film_id = fa2.film_id
+JOIN 
+    sakila.actor a1 ON fa1.actor_id = a1.actor_id
+JOIN 
+    sakila.actor a2 ON fa2.actor_id = a2.actor_id
+WHERE 
+    a1.actor_id < a2.actor_id; 
+    
+-- 2
+SELECT
+    CONCAT(c1.first_name, ' ', c1.last_name) AS 'CUSTOMER 1',
+    CONCAT(c2.first_name, ' ', c2.last_name) AS 'CUSTOMER 2',
+    f.title AS 'FILM',
+    COUNT(*) AS 'RENTAL_COUNT'
+FROM
+    sakila.rental AS r1
+JOIN
+    sakila.rental AS r2 ON r1.inventory_id = r2.inventory_id AND r1.customer_id < r2.customer_id
+JOIN
+    sakila.customer AS c1 ON r1.customer_id = c1.customer_id
+JOIN
+    sakila.customer AS c2 ON r2.customer_id = c2.customer_id
+JOIN
+    sakila.inventory AS i1 ON r1.inventory_id = i1.inventory_id
+JOIN
+    sakila.inventory AS i2 ON r2.inventory_id = i2.inventory_id
+JOIN
+    sakila.film AS f ON i1.film_id = f.film_id
+GROUP BY
+    'CUSTOMER 1', 'CUSTOMER 2', 'FILM'
+HAVING
+    COUNT(*) > 3
+LIMIT 0, 1000;
+
+-- 3
+SELECT
+    a.first_name AS 'Actor First Name',
+    a.last_name AS 'Actor Last Name',
+    f.title AS 'Film Title'
+FROM
+    sakila.actor AS a
+CROSS JOIN
+    sakila.film AS f;