Упростите выражения: \(\frac{2\sin(\alpha+2\pi)+\cos(-\pi/2+\alpha)}{2\sin(\alpha+\pi)}\); \(\frac{\cos(3\pi-\beta)+2\sin(-3\pi/2+\beta)}{5\cos(\beta+2\pi)}\).

1. Используем периодичность и формулы приведения: \(\sin(\alpha+2\pi)=\sin\alpha\), \(\cos(\alpha-\frac{\pi}{2})=\sin\alpha\), \(\sin(\alpha+\pi)=-\sin\alpha\).

\(\frac{2\sin(\alpha+2\pi)+\cos(-\frac{\pi}{2}+\alpha)}{2\sin(\alpha+\pi)}=\frac{2\sin\alpha+\sin\alpha}{-2\sin\alpha}=-\frac{3}{2}\), при \(\sin\alpha\ne0\).

2. \(\cos(3\pi-\beta)=-\cos\beta\), \(\sin(\beta-\frac{3\pi}{2})=\cos\beta\), а \(\cos(\beta+2\pi)=\cos\beta\).

\(\frac{\cos(3\pi-\beta)+2\sin(-\frac{3\pi}{2}+\beta)}{5\cos(\beta+2\pi)}=\frac{-\cos\beta+2\cos\beta}{5\cos\beta}=\frac{1}{5}\), при \(\cos\beta\ne0\).

Ответ: 1) \(-\frac{3}{2}\); 2) \(\frac{1}{5}\).