Ответ:

Используем периодичность и формулы приведения.

  1. \(\operatorname{tg}\frac{5\pi}{4}=\operatorname{tg}\left(\pi+\frac{\pi}{4}\right)=\operatorname{tg}\frac{\pi}{4}=1\).
  2. \(\sin\frac{7\pi}{6}=\sin\left(\pi+\frac{\pi}{6}\right)=-\sin\frac{\pi}{6}=-\frac12\).
  3. \(\cos\frac{5\pi}{3}=\cos\left(2\pi-\frac{\pi}{3}\right)=\cos\frac{\pi}{3}=\frac12\).
  4. \(\operatorname{ctg}\frac{5\pi}{3}=\operatorname{ctg}\left(2\pi-\frac{\pi}{3}\right)=-\operatorname{ctg}\frac{\pi}{3}=-\frac{\sqrt3}{3}\).
  5. \(\sin\left(-\frac{13\pi}{6}\right)=\sin\left(-2\pi-\frac{\pi}{6}\right)=\sin\left(-\frac{\pi}{6}\right)=-\frac12\).
  6. \(\cos\left(-\frac{7\pi}{3}\right)=\cos\left(-2\pi-\frac{\pi}{3}\right)=\cos\left(-\frac{\pi}{3}\right)=\frac12\).
  7. \(\operatorname{tg}\left(-\frac{2\pi}{3}\right)=\operatorname{tg}\left(-\frac{2\pi}{3}+\pi\right)=\operatorname{tg}\frac{\pi}{3}=\sqrt3\).
  8. \(\operatorname{ctg}\left(-\frac{7\pi}{4}\right)=\operatorname{ctg}\left(-\frac{7\pi}{4}+2\pi\right)=\operatorname{ctg}\frac{\pi}{4}=1\).

Ответ: 1) \(1\); 2) \(-\frac12\); 3) \(\frac12\); 4) \(-\frac{\sqrt3}{3}\); 5) \(-\frac12\); 6) \(\frac12\); 7) \(\sqrt3\); 8) \(1\).