Преобразуем выражение:
$$\sqrt{18} cos^2 \frac{7π}{8} - \sqrt{18} sin^2 \frac{7π}{8} = \sqrt{18} (cos^2 \frac{7π}{8} - sin^2 \frac{7π}{8}) = \sqrt{18} cos(\frac{2*7π}{8}) = \sqrt{18} cos(\frac{7π}{4})$$ $$\sqrt{18} cos(\frac{7π}{4}) = \sqrt{18} cos(2π - \frac{π}{4}) = \sqrt{18} cos(\frac{π}{4}) = \sqrt{18} * \frac{\sqrt{2}}{2} = \frac{\sqrt{36}}{2} = \frac{6}{2} = 3$$Ответ: 3