Answer
(a) $ f^{-1}(x)=\frac{1}{x}+2$.
(b) $f$: $(-\infty,2)U(2,\infty)$ and $(-\infty,0)U(0,\infty)$. $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,2)U(2,\infty)$.
(c) See graph.
Work Step by Step
(a) $f(x)=\frac{1}{x-2} \longrightarrow y=\frac{1}{x-2}\longrightarrow x=\frac{1}{y-2}\longrightarrow y=\frac{1}{x}+2\longrightarrow f^{-1}(x)=\frac{1}{x}+2$. Check $f(f^{-1})=x$ and $f^{-1}(f)=x$
(b) Domain and range of $f$: $(-\infty,2)U(2,\infty)$ and $(-\infty,0)U(0,\infty)$. Domain and range of $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,2)U(2,\infty)$.
(c) See graph.
