Ответ:
\[\cos135^\circ=\cos(180^\circ-45^\circ)=-\cos45^\circ=-\frac{\sqrt2}{2}.\]
\[\sin\frac{8\pi}{3}=\sin\left(2\pi+\frac{2\pi}{3}\right)=\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}.\]
\[\operatorname{tg}\frac{7\pi}{3}=\operatorname{tg}\left(2\pi+\frac\pi3\right)=\operatorname{tg}\frac\pi3=\sqrt3.\]
Используем формулу \(\cos^2x-\sin^2x=\cos2x\):
\[\cos^2\frac\pi8-\sin^2\frac\pi8=\cos\frac\pi4=\frac{\sqrt2}{2}.\]
Ответ: 1) \(-\frac{\sqrt2}{2}\); 2) \(\frac{\sqrt3}{2}\); 3) \(\sqrt3\); 4) \(\frac{\sqrt2}{2}\).
