Нужен ход решения!!!!!!!!!!!

Решите задачу:

$1)\sqrt{2}sin\frac{13\pi}{8}cos\frac{13\pi}{8}=\frac{\sqrt2}{2}sin\frac{26\pi}{8}=\frac{1}{\sqrt2}}sin\frac{13\pi}{4}=\frac{1}{\sqrt2}*-\frac{1}{\sqrt2}=-0.5\\\\2)\sqrt{75}cos^2\frac{7\pi}{12}-\sqrt{75}sin^2\frac{7\pi}{12}=\sqrt{75}*cos\frac{14\pi}{12}=5\sqrt{3}*-\frac{\sqrt3}{2}=-7.5\\\\3)\sqrt{32}cos^2\frac{\pi}{8}-\sqrt8=\sqrt8(2cos^2\frac{\pi}{8}-1)=2\sqrt2*cos\frac{\pi}{4}=2\sqrt2*\frac{1}{\sqrt2}=2$

$4)\sqrt{48}-\sqrt{192}sin^2\frac{19\pi}{12}=\sqrt{48}(1-2sin^2\frac{19\pi}{12})=4\sqrt3*cos\frac{38\pi}{12}=\\=4\sqrt3*-\frac{\sqrt3}{2}=-6$