(i) To express \(x^2 - 2x - 15\) in the form \((x + a)^2 + b\), complete the square:

\(x^2 - 2x - 15 = (x-1)^2 - 1 - 15 = (x-1)^2 - 16\).

(ii) The smallest possible value of \(c\) is the minimum value of \((x-1)^2 - 16\), which is \(-16\).

(iii) Given \(c = 9\) and \(d = 65\), solve:

\(9 \leq (x-1)^2 - 16 \leq 65\)

\(25 \leq (x-1)^2 \leq 81\)

\(5 \leq x-1 \leq 9\)

\(6 \leq x \leq 10\)

Thus, \(p = 6\) and \(q = 10\).

(iv) To find \(f^{-1}(x)\), start with \(x = (y-1)^2 - 16\):

\(x + 16 = (y-1)^2\)

\(y - 1 = \pm \sqrt{x + 16}\)

Since \(y \geq 1\), \(y - 1 = \sqrt{x + 16}\)

\(y = 1 + \sqrt{x + 16}\)

Thus, \(f^{-1}(x) = 1 + \sqrt{x + 16}\).