(i) The smallest value of \(c\) is 2. This is because the function \(f(x) = (x-2)^2 + 2\) is defined for \(x \geq c\), and the smallest value for which the function is one-to-one is when \(c = 2\).

(ii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 2\). Rearrange to solve for \(x\):

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

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

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

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

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

The domain of \(f^{-1}\) is \(x > 6\) because \(f(x)\) is defined for \(x \geq 4\) and \(f(4) = 6\).

(iii) Solve \(ff(x) = 51\):

\(f(f(x)) = f((x-2)^2 + 2) = ((x-2)^2 + 2 - 2)^2 + 2 = ((x-2)^2)^2 + 2\)

Set \(((x-2)^2)^2 + 2 = 51\)

\(((x-2)^2)^2 = 49\)

\((x-2)^2 = 7\)

\(x-2 = \pm \sqrt{7}\)

Since \(x \geq 4\), \(x = 2 + \sqrt{7}\).