Ответ:
[tex] arccos0-arctg1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\; ;\\\\\\arcsin(-\frac{1}{2})+arctg\sqrt3=-arcsin\frac{1}{2}+\frac{\pi }{3}=-\frac{\pi }{6}+\frac{\pi }{3}= \frac{\pi }{6}\; ;\\\\\\arccos(-\frac{\sqrt2}{2})-3arcsin(-\frac{1}{2})=\pi -arcsin\frac{\sqrt2}{2} +3arcsin\frac{1}{2}=\\\\=\pi -\frac{\pi }{4}+3\cdot \frac{\pi}{6}=\frac{3\pi }{4}+\frac{\pi }{2}=\frac{5\pi }{4}\; ; [/tex]