(a) To express \(f(x) = -2x^2 - 8x - 13\) in the form \(-2(x + a)^2 + b\), complete the square:

\(f(x) = -2(x^2 + 4x) - 13\)

\(= -2((x + 2)^2 - 4) - 13\)

\(= -2(x + 2)^2 + 8 - 13\)

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

Thus, \(a = 2\) and \(b = -5\).

(b) The vertex form \(-2(x + 2)^2 - 5\) shows the maximum value is \(-5\) when \(x = -2\). Since \(x < -3\), \(f(x) < -7\).

(c) To find \(f^{-1}(x)\), start with \(y = -2(x + 2)^2 - 5\).

Rearrange to solve for \(x\):

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

\((x + 2)^2 = -\frac{y + 5}{2}\)

\(x + 2 = \pm \sqrt{-\frac{y + 5}{2}}\)

\(x = -2 \pm \sqrt{-\frac{y + 5}{2}}\)

Thus, \(f^{-1}(x) = -2 - \sqrt{\frac{x+5}{2}}\) or \(f^{-1}(x) = -2 + \sqrt{\frac{x+5}{2}}\).