Proving that activating inverse function on a group intersection is equal to the intersection of inverse activated on groups

Question

I have been trying to figure out a direction to this question, but to no avail:

Let $f:A\rightarrow B$

$C_1,C_2\subseteq B$

Prove that:

1. $f^{-1}\left(C_{1}\cap C_{2}\right)=f^{-1}\left(C_{1}) \cap f^{-1}(C_{2}\right)$.

2. $f^{-1}\left(C_{1}\cup C_{2}\right)=f^{-1}\left(C_{1}) \cup f^{-1}(C_{2}\right)$.

Answer 1

x in f$^{-1}$(A $\cap$ B) iff f(x) in A $\cap$ B
iff f(x) in A and f(x) in B iff ...