given: A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To prove: A × C ⊂ B × D

LHS: A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

RHS: B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Since, all elements of A × C is in B × D.

∴We can A × C ⊂ B × D

(ii) To prove: A × (B ∩ C) = (A × B) ∩ (A × C)

LHS: (B ∩ C) = ∅

A × (B ∩ C) = ∅

RHS: (A × B) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

(A × C) = {(1, 5), (1, 6), (2, 5), (2, 6)}

Since, there is no common element between A × B and A × C

(A × B) ∩ (A × C) = ∅

∴ LHS = RHS