(i) To express \(2x^2 - 12x + 7\) in the form \(2(x + a)^2 + b\), complete the square:

\(2x^2 - 12x + 7 = 2(x^2 - 6x) + 7\)

\(= 2((x - 3)^2 - 9) + 7\)

\(= 2(x - 3)^2 - 18 + 7\)

\(= 2(x - 3)^2 - 11\)

Thus, \(a = -3\) and \(b = -11\).

(ii) The range of \(f\) is \(y > -11\) because the vertex of the parabola is at its minimum point.

(iii) For \(g\) to have an inverse, it must be one-to-one. The function is one-to-one for \(x \leq 3\), so the largest value of \(k\) is \(k < 3\).

(iv) To find \(g^{-1}(x)\), start with:

\(y = 2(x - 3)^2 - 11\)

\(y + 11 = 2(x - 3)^2\)

\(\frac{y + 11}{2} = (x - 3)^2\)

\(x - 3 = \pm \sqrt{\frac{y + 11}{2}}\)

Since \(x \leq 3\), choose the negative root:

\(x = 3 - \sqrt{\frac{y + 11}{2}}\)

Thus, \(g^{-1}(x) = 3 - \sqrt{\frac{x + 11}{2}}\).