(i) To express \(-2x^2 + 12x - 3\) in the form \(-2(x+a)^2 + b\), we complete the square:

\(-2x^2 + 12x - 3 = -2(x^2 - 6x) - 3\)

Complete the square for \(x^2 - 6x\):

\(x^2 - 6x = (x-3)^2 - 9\)

Substitute back:

\(-2((x-3)^2 - 9) - 3 = -2(x-3)^2 + 18 - 3 = -2(x-3)^2 + 15\)

Thus, \(a = -3\) and \(b = 15\).

(ii) The greatest value of \(f(x)\) is the maximum value of \(-2(x-3)^2 + 15\), which occurs when \((x-3)^2 = 0\). Therefore, the greatest value is 15.

(iii) To find \(x\) for which \(gf(x) + 1 = 0\), first find \(gf(x)\):

\(gf(x) = g(f(x)) = 2(-2x^2 + 12x - 3) + 5 = -4x^2 + 24x - 6 + 5 = -4x^2 + 24x - 1\)

Set \(gf(x) + 1 = 0\):

\(-4x^2 + 24x - 1 + 1 = 0\)

\(-4x^2 + 24x = 0\)

Factor out \(-4x\):

\(-4x(x - 6) = 0\)

Thus, \(x = 0\) or \(x = 6\).