Ответ:

  1. \(\operatorname{arctg}\left(-\frac{\sqrt3}{3}\right)=-\frac{\pi}{6}\), поскольку \(\operatorname{tg}\left(-\frac{\pi}{6}\right)=-\frac{\sqrt3}{3}\).
  2. \(\arccos\frac{\sqrt2}{2}=\frac{\pi}{4}\).
  3. \(\operatorname{arctg}1+\arccos\frac{\sqrt2}{2}=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}\).
  4. \(\operatorname{ctg}\left(\frac{\pi}{2}-\arccos\frac{\sqrt2}{2}\right)=\operatorname{ctg}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\operatorname{ctg}\frac{\pi}{4}=1\).

Ответ: а) \(-\frac{\pi}{6}\); б) \(\frac{\pi}{4}\); в) \(\frac{\pi}{2}\); г) \(1\).