(i) The function \(f(x) = 2x + 3\). To find \(ff(x)\), we substitute \(f(x)\) into itself: \(f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9\). Set \(4x + 9 = 25\), solving gives \(x = 4\). Alternatively, \(f(11) = 25\) and \(f(4) = 11\).

(ii) Set \(f(x) = g(x)\): \(2x + k = x^2 - 6x + 8\). Rearrange to \(x^2 - 8x + 8 - k = 0\). For no real solutions, the discriminant \(b^2 - 4ac < 0\). Here, \(a = 1, b = -8, c = 8-k\). So, \((-8)^2 - 4(1)(8-k) < 0\) simplifies to \(64 - 32 + 4k < 0\), giving \(k < -8\).

(iii) Start with \(y = x^2 - 6x + 8\). Complete the square: \(y = (x-3)^2 - 1\). Solve for \(x\): \(y + 1 = (x-3)^2\), \(x - 3 = \pm \sqrt{y + 1}\). Since \(x > 3\), choose the positive root: \(x = \sqrt{y + 1} + 3\). Thus, \(h^{-1}(x) = \sqrt{x + 1} + 3\).