\parstyle\begin{eqnarray*}
&Pf)& x \in f^{-1}(A\cup B)\\
&\Leftrightarrow& f(x) \in A\cup B\\
&\Leftrightarrow& f(x) \in A\mbox{ or }f(x) \in B\\
&\Leftrightarrow& x \in f^{-1}(A)\mbox{ or }x \in f^{-1}(B)\\
&\Leftrightarrow& x \in f^{-1}(A)\cup f^{-1}(B)
\end{eqnarray*}

\parstyle\begin{eqnarray*}
&Pf)& x \in f^{-1}(A\cap B)\\
&\Leftrightarrow& f(x) \in A\cap B\\
&\Leftrightarrow& f(x) \in A\mbox{ and }f(x) \in B\\
&\Leftrightarrow& x \in f^{-1}(A)\mbox{ and }x \in f^{-1}(B)\\
&\Leftrightarrow& x \in f^{-1}(A)\cap f^{-1}(B)
\end{eqnarray*}

\parstyle\begin{eqnarray*}
&&Pf) (\Rightarrow)\\
&&\mbox{Let }y \in f(A\cup B).\mbox{ Then }y=f(x)\mbox{ where }x \in A\cup B.\\
&&y=f(x)\mbox{ where }x \in A\mbox{ or }x \in B.\mbox{ So }y\in f(A)\mbox{ or }y\in f(B).\\
&&\therefore y\in f(A)\cup f(B).\\
&&(\Leftarrow)\\
&&\mbox{Let }y\in f(A)\cup f(B).\mbox{ Then }y\in f(A)\mbox{ or }y\in f(B).\\
&&y=f(a)\mbox{ for some }a \in A\mbox{ or }y=f(b)\mbox{ for some }b \in B.\\
&&\mbox{So }y=f(x)\mbox{ for some }x \in A\cup B.\\
&&\therefore y \in f(A\cup B).
\end{eqnarray*}

\parstyle\begin{eqnarray*}
&&\mbox{counter example)}\\
&&\mbox{Let }f:X\rightarrow Y\mbox{ be a ftn that is not 1-1.}\\
&&\mbox{Then choose distinct }a, b \in X\mbox{ such that } f(a)=f(b)=y.\\
&&\mbox{And let }A, B\mbox{ be two sets satisfying }A\cap B=\emptyset\mbox{ and }a\in A, b\in B.\\
&&\mbox{Then }y\in f(A) \cap f(B)\mbox{ but }f(A\cap B)=\emptyset.
\end{eqnarray*}