г) ctg (2arcsin (-)) + sin(3arccos(-)) д) 2arccos (2) + arctg(-1) + 2arcsin (-)

г) $$\cot(2\arcsin(-\frac{1}{2})) + \sin(3\arccos(-\frac{\sqrt{3}}{2}))$$. $$\arcsin(-\frac{1}{2}) = -\frac{\pi}{6}$$, $$\arccos(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6}$$. Тогда $$\cot(2 \cdot (-\frac{\pi}{6})) + \sin(3 \cdot \frac{5\pi}{6}) = \cot(-\frac{\pi}{3}) + \sin(\frac{5\pi}{2}) = -\frac{\sqrt{3}}{3} + \sin(2\pi + \frac{\pi}{2}) = -\frac{\sqrt{3}}{3} + 1$$

д) $$2\arccos(\frac{\sqrt{2}}{2}) + \arctan(-1) + 2\arcsin(-\frac{\sqrt{3}}{2})$$. $$\arccos(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$$, $$\arctan(-1) = -\frac{\pi}{4}$$, $$\arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}$$. Тогда $$2 \cdot \frac{\pi}{4} - \frac{\pi}{4} + 2 \cdot (-\frac{\pi}{3}) = \frac{\pi}{2} - \frac{\pi}{4} - \frac{2\pi}{3} = \frac{6\pi - 3\pi - 8\pi}{12} = -\frac{5\pi}{12}$$

Ответ: г) $$1 - \frac{\sqrt{3}}{3}$$, д) $$- \frac{5\pi}{12}$$