(a) The function \(f(x) = (x - 2)^2 - 4\) is a quadratic function with vertex at \((2, -4)\). Since it opens upwards, the minimum value is \(-4\). Thus, the range is \(f(x) > -4\).

(b) To find \(f^{-1}(x)\), set \(y = (x - 2)^2 - 4\) and solve for \(x\):

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

\(x - 2 = \pm \sqrt{y + 4}\)

Since \(x \geq 2\), we take the positive root: \(x = \sqrt{y + 4} + 2\)

Thus, \(f^{-1}(x) = \sqrt{x + 4} + 2\).

(c) Given \(a = -\frac{5}{3}\), solve \(f(x) = g(x)\):

\((x - 2)^2 - 4 = -\frac{5}{3}x + 2\)

\(x^2 - 4x + 4 - 4 = -\frac{5}{3}x + 2\)

\(x^2 - 4x + \frac{5}{3}x - 2 = 0\)

\((3x + 2)(x - 3) = 0\)

\(x = 3\) only.

(d) Given \(gg f^{-1}(12) = 62\):

\(f^{-1}(12) = 6\)

\(g(f^{-1}(12)) = 6a + 2\)

\(g(g(f^{-1}(12))) = a(6a + 2) + 2 = 62\)

\(6a^2 + 2a - 60 = 0\)

Solving the quadratic equation gives \(a = \frac{10}{3}\) or \(a = 3\).