Answer
See below.
Work Step by Step
$f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}$
$f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}+0$
$f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}+[\frac{f(-x)}{2}-\frac{f(-x)}{2}]$
$f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}+\frac{f(-x)}{2}-\frac{f(-x)}{2}$
... addition of real numbers is commutative
$f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(-x)}{2}+\frac{f(x)}{2}-\frac{f(-x)}{2}$
... and associative
$f(x)=[\displaystyle \frac{f(x)}{2}+\frac{f(-x)}{2}]+[\frac{f(x)}{2}-\frac{f(-x)}{2}]$
$f(x)=\displaystyle \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$
... which we needed to show
$F_{1}(x)=\displaystyle \frac{f(x)+f(-x)}{2}\qquad $ is even because
$F_{1}(-x)=\displaystyle \frac{f(-x)+f(x)}{2}=\frac{f(x)+f(-x)}{2}=F_{1}(x)$
$F_{1}(-x)=F_{1}(x)$
$F_{2}(x)=\displaystyle \frac{f(x)-f(-x)}{2}\qquad $ is odd because
$F_{2}(-x)=\displaystyle \frac{f(-x)-f(x)}{2}=\frac{-[f(x)-f(-x)]}{2}=-F_{2}(x)$
$F_{2}(-x)=F_{2}(x)$
