a) (A ∩ B) ⊆ A

Let x be in (A ∩ B), then x is in A and x is in B. Hence x is in A. Thus (A ∩ B) ⊆ A

b) A ⊆ A ∪ B

This one seems so easy to prove, I'm not sure what else to say about it:

Let x be in A. If x is in A then x is in A ∪ B. Thus A ⊆ A ∪ B

c) A ∩ (B - A) = 0

(B - A) = {x | x is in B and x is not in A}. So intuitively this equality should make sense. You take everything that is in A and intersect with everything that is in B takeaway A, so A and (B-A) should be mutually disjoint sets).

For some reason I can't think of how to prove this one, sorry. Someone else will have to tackle it.