(a) To express \(-3x^2 + 12x + 2\) in the form \(-3(x-a)^2 + b\), complete the square:

\(-3(x^2 - 4x) + 2\)

\(-3((x-2)^2 - 4) + 2\)

\(-3(x-2)^2 + 12 + 2\)

\(-3(x-2)^2 + 14\)

(b) The vertex form \(-3(x-2)^2 + 14\) shows the maximum value is 14 when \(x = 2\). Thus, the largest possible value of \(k\) is 2.

(c) Given \(k = -1\), substitute into \(-3(x-2)^2 + 14\) to find the range:

\(-3((-1)-2)^2 + 14 = -3(9) + 14 = -27 + 14 = -13\)

Thus, the range is \(y \leq -13\).

(d) To find \(f^{-1}(x)\), start from \(y = -3(x-2)^2 + 14\):

\((x-2)^2 = \frac{14-y}{3}\)

\(x-2 = \pm \sqrt{\frac{14-y}{3}}\)

Since \(x \leq 2\), take the negative root:

\(x = 2 - \sqrt{\frac{14-y}{3}}\)

Thus, \(f^{-1}(x) = 2 - \sqrt{\frac{14-x}{3}}\).