Computing the composition of a piecewise function Question:
$$ \text{Define } f:\mathbb{Z}\rightarrow\mathbb{Z}\text{ by }f(x)=\begin{cases}x+3\text{ if }x\text{ is ODD}\\ x-5\text{ if }x\text{ is EVEN}\end{cases} $$
Compute $ \  f \circ f$
My attempt:
$ \  (f \circ f)(x) = f(f(x))$
From here do I have to create another piecewise function and consider the cases when $ f(x) = x+3$ and when $ f(x) = x-5$? 

Yes, you could regard the composition as a piecewise function.
$$(f\circ f)(n) = f( f(n) ) = \begin{cases}
f(n) + 3 & \text{if $f(n)$ is odd, and} \\
f(n)-5 & \text{if $f(n)$ is even}. \\
\end{cases}$$
But notice that $f(n)$ is $n$, plus some odd integer (either $+3$ or $-5$).  Thus $f(n)$ is even when $n$ is odd, and vice versa.  Thus we have
$$f( f(n) ) = \begin{cases}
f(n) + 3 & \text{if $n$ is even, and} \\
f(n)-5 & \text{if $n$ is odd}. \\
\end{cases}$$
But then we know what to do with $f(n)$.  When $n$ is even, $f(n) = n-5$, so $$f(n) + 3 = (n-5) + 3 = n-2.$$  Thus the first part of the piecewise definition can be rewritten to give
$$f( f(n) ) = \begin{cases}
n-2 & \text{if $n$ is even, and} \\
f(n)-5 & \text{if $n$ is odd}. \\
\end{cases}$$
Similarly, of $n$ is odd, then $f(n) = n+3$, and so
$$f(n)-5 = (n+3)-5 = n-2.$$
Hence the second part of the piecewise defined composition can be rewritten, giving
$$f( f(n) ) = \begin{cases}
n-2 & \text{if $n$ is even, and} \\
n-2 & \text{if $n$ is odd}. \\
\end{cases}$$
Since we get the same thing either way, we conclude that
$$(f\circ f)(n) = n-2.$$

That begin said, it is perhaps more direct to do the following:  there are two cases to consider, either $n$ is even, or $n$ is odd (since all integers are either odd or even).  If $n$ is even, then
$$(f\circ f)(n) = f(f(n)) = f(n-5).$$
But $n-5$ is odd (even - odd = odd), so
$$f(n-5) = (n-5) + 3 = n-2.$$
Similarly, if $n$ is odd, then
$$(f\circ f)(n) = f(n+3) = (n+3) - 5 = n-2.$$
In either case, we conclude that
$$(f\circ f)(n) = n-2.$$
