(i) To solve \(ff(x) = 11\), first find \(f(f(x))\):

\(f(x) = 2x - 3\)

\(f(f(x)) = f(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\)

Set \(4x - 9 = 11\):

\(4x = 20\)

\(x = 5\)

(ii) The function \(g(x) = x^2 + 4x\) is a quadratic function. The vertex form is \(g(x) = (x + 2)^2 - 4\), with a minimum at \(x = -2\).

The minimum value is \(-4\), so the range is \(g(x) \geq -4\).

(iii) To find where \(g(x) > 12\):

\(x^2 + 4x > 12\)

\(x^2 + 4x - 12 > 0\)

Factor the quadratic: \((x - 2)(x + 6) > 0\)

The critical points are \(x = 2\) and \(x = -6\).

Test intervals: \(x < -6\), \(-6 < x < 2\), \(x > 2\).

The solution is \(x < -6\) or \(x > 2\).