this is pretty straightforward if you just apply the definitions.
The general way to show two sets are equal is to show they are both subsets of one another, i.e. $A \subset B \wedge B \subset A \Rightarrow A=B$
I'll do one direction and leave the other to you.
let $(x,y) \in (A \times B) \cap (C \times D)$
then by definition $(x,y) \in (A \times B) \wedge (x,y) \in (C \times D)$
then again by definition this implies
$x \in A \wedge x \in C,~~y \in B \wedge y \in D$
thus $x \in A \cap C, ~~y \in B \cap D$ and thus $(x,y) \in (A \cap C) \times (B \cap D)$
thus $ (A \times B) \cap (C \times D) \subset (A \cap C) \times (B \cap D)$
now you do the other direction.
Am I wrong, is the first question not, Prove that $\displaystyle A\times (B\setminus C)=(A\times B)\setminus (A\times C)\Large\color{red}??$
$\displaystyle \begin{align*}(a,d)\in (A\times (B\setminus C)&\iff (a\in A) \wedge [d\in (B\setminus C)]\\&\iff (a\in A) \wedge[d\in B\wedge d\notin C]\\&\iff [(a\in A)\wedge (d\in B)] \wedge[d\in A\wedge d\notin C] \\&\iff (a,d)\in [A\times B]\setminus [A\times C]\end{align*}$