Ответ:

  1. Применяем формулу \(\cos^2x-\sin^2x=\cos2x\):

    \(7\sqrt2\left(\cos^2\frac{5\pi}{8}-\sin^2\frac{5\pi}{8}\right)=7\sqrt2\cos\frac{5\pi}{4}=7\sqrt2\cdot\left(-\frac{\sqrt2}{2}\right)=-7\).

  2. \(\cos^2\frac{17\pi}{12}=\cos^2\frac{5\pi}{12}=\frac{2-\sqrt3}{4}\).

    \(\sqrt{48}\cdot\frac{2-\sqrt3}{4}-\sqrt{12}=4\sqrt3\cdot\frac{2-\sqrt3}{4}-2\sqrt3=\sqrt3(2-\sqrt3)-2\sqrt3=-3\).

  3. \(\sqrt{50}=5\sqrt2\), \(\sqrt{200}=10\sqrt2\), а \(\sin^2\frac{5\pi}{8}=\frac{2+\sqrt2}{4}\).

    \(5\sqrt2-10\sqrt2\cdot\frac{2+\sqrt2}{4}=5\sqrt2-(5\sqrt2+5)=-5\).

Ответ: 1) −7; 2) −3; 3) −5.