(a) The function \(f(x) = -3x^2 + 2\) is a downward-opening parabola. The vertex occurs at \(x = 0\), but since \(x \leq -1\), the maximum value of \(f(x)\) is at \(x = -1\). Calculating \(f(-1) = -3(-1)^2 + 2 = -1\). Therefore, the range of \(f\) is \(y \leq -1\).

(b) To find the inverse, start with \(y = -3x^2 + 2\). Rearrange to solve for \(x\):

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

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

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

Since \(x \leq -1\), we take the negative root:

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

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