(i) To find \(f^{-1}(x)\), start with \(y = 2x + 3\).

Solve for \(x\):

\(y - 3 = 2x\)

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

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

To solve \(f(x) = f^{-1}(x)\):

\(2x + 3 = \frac{1}{2}x - \frac{3}{2}\)

Multiply through by 2 to clear the fraction:

\(4x + 6 = x - 3\)

\(3x = -9\)

\(x = -3\)

(iii) Find \(gf(x) \leq 16\):

\(gf(x) = g(2x + 3) = (2x + 3)^2 - 6(2x + 3)\)

\(= 4x^2 + 12x + 9 - 12x - 18\)

\(= 4x^2 - 9\)

Set \(4x^2 - 9 \leq 16\):

\(4x^2 \leq 25\)

\(x^2 \leq \frac{25}{4}\)

\(-\frac{5}{2} \leq x \leq \frac{5}{2}\)

Given \(x \leq 0\), the solution is \(-\frac{5}{2} \leq x \leq 0\).