13. Given the function
$$
\begin{aligned}
f(x)= & \sin \left(x+\frac{\pi}{4}\right)+2 \sin \left(x-\frac{\pi}{4}\right)- \\
& 4 \cos 2 x+3 \cos \left(x+\frac{3 \pi}{4}\right) .
\end{aligned}
$$
(1) Determine whether the function $f(x)$ is odd, even, or neither, and provide a proof;
(2) Find the minimum and maximum values of $f(x)$ on $\left[\frac{\pi}{2}, \pi\right]$.
Three, 13. (1) Notice that
$$
\begin{aligned}
f(x)= & \frac{\sqrt{2}}{2}(\sin x+\cos x)+\sqrt{2}(\sin x-\cos x)- \\
& 4 \cos 2 x-\frac{3 \sqrt{2}}{2}(\cos x+\sin x) \\
= & -2 \sqrt{2} \cos x-4 \cos 2 x .
\end{aligned}
$$

Since $\cos x$ and $\cos 2 x$ are both even functions, $f(x)$ is an even function.
(2) Notice that
$$
\begin{array}{l}
f(x)=-2 \sqrt{2} \cos x-4\left(2 \cos ^{2} x-1\right) \\
=-8 \cos ^{2} x-2 \sqrt{2} \cos x+4 \\
=-8\left(\cos x+\frac{\sqrt{2}}{8}\right)^{2}+\frac{17}{4} .
\end{array}
$$

Given $x \in\left[\frac{\pi}{2}, \pi\right]$, we know $-1 \leqslant \cos x \leqslant 0$.
Thus, when $\cos x=-1$, $(f(x))_{\min }=2 \sqrt{2}-4$;
when $\cos x=-\frac{\sqrt{2}}{8}$, $(f(x))_{\max }=\frac{17}{4}$.