(a) The function \(f(x) = 2 - \frac{5}{x+2}\) has a horizontal asymptote at \(y = 2\). As \(x \to \infty\), \(f(x) \to 2\). Therefore, the range of \(f\) is \(y < 2\).

(b) To find the inverse, start with \(y = 2 - \frac{5}{x+2}\).

Rearrange to solve for \(x\):

\(y(x+2) = 2(x+2) - 5\)

\(yx + 2y = 2x + 4 - 5\)

\(yx + 2y = 2x - 1\)

\(2y + 1 = 2x - yx\)

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

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

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

The domain of \(f^{-1}\) is determined by the range of \(f\), which is \(x < 2\).