Answer
(a) $ f^{-1}(x)=\frac{2}{\sqrt {1-2x}}$.
(b) $f$: $(0,\infty)$ and $(-\infty,\frac{1}{2})$.
$f^{-1}$: $(-\infty,\frac{1}{2})$ and $(0,\infty)$.
Work Step by Step
(a) $f(x)=\frac{x^2-4}{2x^2}, x\gt0\longrightarrow y=\frac{x^2-4}{2x^2} \longrightarrow x=\frac{y^2-4}{2y^2}, y\gt0 \longrightarrow y=\frac{2}{\sqrt {1-2x}} \longrightarrow f^{-1}(x)=\frac{2}{\sqrt {1-2x}}$. Check $f(f^{-1})=\frac{(\frac{2}{\sqrt {1-2x}})^2-4}{2(\frac{2}{\sqrt {1-2x}})^2}=x$ and $f^{-1}(f)=\frac{2}{\sqrt {1-2(\frac{x^2-4}{2x^2})}}=x$
(b) Domain and range of $f$: $(0,\infty)$ and $(-\infty,\frac{1}{2})$. Domain and range of $f^{-1}$: $(-\infty,\frac{1}{2})$ and $(0,\infty)$.