б) Используем формулу разности косинусов:
$$cos x - cos y = -2 sin(\frac{x+y}{2}) sin(\frac{x-y}{2})$$
В нашем случае $$x = \frac{5\pi}{12}$$ и $$y = \frac{\pi}{12}$$:
$$cos \frac{5\pi}{12} - cos \frac{\pi}{12} = -2 sin(\frac{\frac{5\pi}{12} + \frac{\pi}{12}}{2}) sin(\frac{\frac{5\pi}{12} - \frac{\pi}{12}}{2}) = -2 sin(\frac{\frac{6\pi}{12}}{2}) sin(\frac{\frac{4\pi}{12}}{2}) = -2 sin(\frac{\pi}{4}) sin(\frac{\pi}{6})$$
$$-2 sin(\frac{\pi}{4}) sin(\frac{\pi}{6}) = -2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = -\frac{\sqrt{2}}{2}$$
Ответ: $$-\frac{\sqrt{2}}{2}$$