化简

(1) $\frac{\cos (\alpha-\pi)}{\sin (\pi-\alpha)} \cdot \sin \left(\alpha-\frac{\pi}{2}\right) \cos \left(\frac{\pi}{2}+\alpha\right) ;$

(2) $\sin (-\alpha-5 \pi) \cos \left(\alpha-\frac{\pi}{2}\right)-\sin \left(\frac{3 \pi}{2}+\alpha\right) \cos (\alpha-2 \pi)$
(1)原式 $=\frac{\cos [-(\pi-\alpha)]}{\sin \alpha} \cdot \sin \left[-\left(\frac{\pi}{2}-\alpha\right)\right](-\sin \alpha)$

$=\frac{\cos (\pi-\alpha)}{\sin \alpha} \cdot\left[-\sin \left(\frac{\pi}{2}-\alpha\right)\right](-\sin \alpha)$

$=\frac{-\cos \alpha}{\sin \alpha} \cdot(-\cos \alpha)(-\sin \alpha)$

$=-\cos ^{2} \alpha$.

(2)原式 $=\sin (-\alpha-\pi) \cos \left[-\left(\frac{\pi}{2}-\alpha\right)\right]+\cos \alpha \cos [-(2 \pi-\alpha)]$

$=\sin [-(\alpha+\pi)] \cos \left(\frac{\pi}{2}-\alpha\right)+\cos \alpha \cos (2 \pi-\alpha)$

$$
\begin{aligned}
& =-\sin (\alpha+\pi) \sin \alpha+\cos \alpha \cos \alpha \\
& =\sin ^{2} \alpha+\cos ^{2} \alpha \\
& =1
\end{aligned}
$$