(i) To express \(9x^2 - 6x + 6\) in the form \((ax + b)^2 + c\), complete the square:

\(9x^2 - 6x + 6 = 9(x^2 - \frac{2}{3}x) + 6\)

\(= 9((x - \frac{1}{3})^2 - \frac{1}{9}) + 6\)

\(= 9(x - \frac{1}{3})^2 - 1 + 6\)

\(= 9(x - \frac{1}{3})^2 + 5\)

Thus, \(a = 3\), \(b = -1\), \(c = 5\).

(ii) The function \(f(x) = 9x^2 - 6x + 6\) is a quadratic function that opens upwards. For \(f\) to be one-one, it must be restricted to the domain where it is either increasing or decreasing. The vertex of the parabola is at \(x = \frac{1}{3}\). Therefore, the smallest value of \(p\) is \(\frac{1}{3}\).

(iii) For \(p = \frac{1}{3}\), the function is one-one for \(x \geq \frac{1}{3}\). To find \(f^{-1}(x)\), start with \(y = (3x-1)^2 + 5\):

\(y - 5 = (3x-1)^2\)

\(\sqrt{y-5} = \pm (3x-1)\)

Since \(x \geq \frac{1}{3}\), we take the positive root:

\(3x - 1 = \sqrt{y-5}\)

\(3x = \sqrt{y-5} + 1\)

\(x = \frac{1}{3}\sqrt{y-5} + \frac{1}{3}\)

Thus, \(f^{-1}(x) = \frac{1}{3}\sqrt{x-5} + \frac{1}{3}\) with domain \(x \geq 5\).

(iv) The equation \(f(x) = q\) has no solution when \(q < 5\) because the minimum value of \(f(x)\) is 5, which occurs at the vertex \(x = \frac{1}{3}\).