Answer
(a) $\frac{\sqrt 3}{2}$, $(\frac{\pi}{5},\frac{\sqrt 3}{2})$
(b) $(\frac{\sqrt 3}{2}, \frac{\pi}{6})$
(c) $(\frac{\pi}{6}, 2)$
Work Step by Step
(a) Given $g(x)=cos(x)$, we have $g(\frac{\pi}{6})=cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$ and point $(\frac{\pi}{6},\frac{\sqrt 3}{2})$
(b) Considering that $g^{-1}$ and $g$ have a symmetry with respect to $y=x$, we can find point $(\frac{\sqrt 3}{2}, \frac{\pi}{6})$ on $g^{-1}$.
(c) Evaluate the function to get $y=2cos(\frac{\pi}{6}-\frac{\pi}{6})=2$, thus the point is $(\frac{\pi}{6}, 2)$
