Решение:

1. $$2 \arcsin \frac{\sqrt{3}}{2} + 3 \arcsin \left(-\frac{1}{2}\right) = 2 \cdot \frac{\pi}{3} + 3 \cdot \left(-\frac{\pi}{6}\right) = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{4\pi - 3\pi}{6} = \frac{\pi}{6}$$
2. $$\arcsin \frac{1}{\sqrt{2}} - 4 \arcsin 1 = \frac{\pi}{4} - 4 \cdot \frac{\pi}{2} = \frac{\pi}{4} - 2\pi = \frac{\pi - 8\pi}{4} = -\frac{7\pi}{4}$$
3. $$\arccos \left(-\frac{1}{2}\right) - \arcsin \frac{\sqrt{3}}{2} = \frac{2\pi}{3} - \frac{\pi}{3} = \frac{\pi}{3}$$
4. $$\arccos (-1) - \arcsin (-1) = \pi - \left(-\frac{\pi}{2}\right) = \pi + \frac{\pi}{2} = \frac{3\pi}{2}$$
5. $$2 \arctan 1 + 3 \arctan \left(-\frac{1}{\sqrt{3}}\right) = 2 \cdot \frac{\pi}{4} + 3 \cdot \left(-\frac{\pi}{6}\right) = \frac{\pi}{2} - \frac{\pi}{2} = 0$$
6. $$4 \arctan (-1) + 3 \arctan \sqrt{3} = 4 \cdot \left(-\frac{\pi}{4}\right) + 3 \cdot \frac{\pi}{3} = -\pi + \pi = 0$$

Ответ:

1. $$\frac{\pi}{6}$$
2. $$-\frac{7\pi}{4}$$
3. $$\frac{\pi}{3}$$
4. $$\frac{3\pi}{2}$$
5. $$0$$
6. $$0$$