(i) To express \(f(x) = 2x^2 - 12x + 13\) in the form \(a(x+b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 6x) + 13\)

\(= 2((x-3)^2 - 9) + 13\)

\(= 2(x-3)^2 - 18 + 13\)

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

Thus, \(a = 2\), \(b = -3\), \(c = -5\).

(ii) The line of symmetry is at \(x = 3\). Since \(0 \leq x \leq A\), \(A = 6\) for symmetry about \(x = 3\).

(iii) The vertex form \(2(x-3)^2 - 5\) shows the minimum value is \(-5\) at \(x = 3\). The maximum value is \(f(0) = 13\). Thus, the range is \(-5 \leq y \leq 13\).

(iv) The function \(g\) is defined for \(x \geq 4\), which is a one-to-one function because it is strictly increasing (as \(4 > 3\)).

(v) To find \(g^{-1}(x)\), start with \(y = 2x^2 - 12x + 13\) and solve for \(x\):

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

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

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

\(x-3 = \pm \sqrt{\frac{y+5}{2}}\)

Since \(x \geq 4\), take the positive root:

\(x = \sqrt{\frac{y+5}{2}} + 3\)

Thus, \(g^{-1}(x) = \sqrt{\frac{x+5}{2}} + 3\).