2) \(\frac13\arctg\sqrt3+\frac12\arctg\left(-\frac{\sqrt3}{3}\right)\)

\(\arctg\sqrt3=\pi/3\), а \(\arctg(-\sqrt3/3)=-\pi/6\).

\[\frac13\cdot\frac{\pi}{3}+\frac12\cdot\left(-\frac{\pi}{6}\right)=\frac{\pi}{9}-\frac{\pi}{12}=\frac{\pi}{36}.\]

Ответ: \(\frac{\pi}{36}\).