(a) The function \(f(x) = 1 + \frac{3}{x-2}\) has a vertical asymptote at \(x = 2\) and a horizontal asymptote at \(y = 1\). As \(x \to \infty\), \(f(x) \to 1\). Therefore, the range of \(f\) is \(y > 1\).

(b) To find \(f^{-1}(x)\), start with \(y = 1 + \frac{3}{x-2}\).

Rearrange to find \(x\):

\(y - 1 = \frac{3}{x-2}\)

\((x-2)(y-1) = 3\)

\(x-2 = \frac{3}{y-1}\)

\(x = \frac{3}{y-1} + 2\)

Thus, \(f^{-1}(x) = \frac{3}{x-1} + 2\).

The domain of \(f^{-1}\) is \(x > 1\) because the original range of \(f\) was \(y > 1\).