Answer
(a) $ f^{-1}(x)=\frac{2}{x}-3$.
(b) $f$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,0)U(0,\infty)$. $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,-3)U(-3,\infty)$.
Work Step by Step
(a) $f(x)=\frac{2}{3+x} \longrightarrow y=\frac{2}{3+x}\longrightarrow x=\frac{2}{3+y}\longrightarrow y=\frac{2}{x}-3\longrightarrow f^{-1}(x)=\frac{2}{x}-3$. Check $f(f^{-1})=\frac{2}{3+\frac{2}{x}-3}=x$ and $f^{-1}(f)=\frac{2}{\frac{2}{3+x}}-3=x$
(b) Domain and range of $f$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,0)U(0,\infty)$. Domain and range of $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,-3)U(-3,\infty)$.