Answer
(a) $\frac{\sqrt 2}{2}$, $(\frac{\pi}{4},\frac{\sqrt 2}{2})$
(b) $(\frac{\sqrt 2}{2}, \frac{\pi}{4})$
(c) $(\frac{\pi}{4}, -2)$
Work Step by Step
(a) Given $f(x)=sin(x)$, we have $f(\frac{\pi}{4})=sin(\frac{\pi}{4})=\frac{\sqrt 2}{2}$ and point $(\frac{\pi}{4},\frac{\sqrt 2}{2})$
(b) Considering that $f^{-1}$ and $f$ have a symmetry with respect to $y=x$, we can find point $(\frac{\sqrt 2}{2}, \frac{\pi}{4})$ on $f^{-1}$.
(c) Evaluate the function to get $y=sin(\frac{\pi}{4}+\frac{\pi}{4})-3=-2$, thus the point is $(\frac{\pi}{4}, -2)$
