Решить первое задание.

$\displaystyle \cos (\pi-\alpha)=\cos \pi \cos \alpha +\sin \pi \sin \alpha=-\cos \alpha\\\\\cos\left( \frac{\pi}{2}-\alpha \right)=\sin \alpha\\\\\sin (2\pi -\alpha)=\sin (-\alpha+2\pi)=\sin (-\alpha)=-\sin \alpha\\\\\sin \left( \frac{3\pi}{2}-\alpha \right)=\sin \frac{3\pi}{2}\cos \alpha-\cos \frac{3\pi}{2}\sin \alpha=-\cos \alpha$

Т.е.

$\displaystyle\frac{-\cos \alpha+\sin \alpha}{-\sin \alpha+\cos \alpha} = \frac{-\cos \alpha+\sin \alpha}{-(-\cos \alpha+\sin \alpha)} =-1$