(ii) To express \(g(x) = 2x^2 - 6x + 5\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 6x + 5 = 2(x^2 - 3x) + 5\)

\(= 2((x - \frac{3}{2})^2 - \frac{9}{4}) + 5\)

\(= 2(x - \frac{3}{2})^2 - \frac{9}{2} + 5\)

\(= 2(x - \frac{3}{2})^2 + \frac{1}{2}\)

Thus, \(a = 2, b = -\frac{3}{2}, c = \frac{1}{2}\).

(iii) The vertex form \(2(x - \frac{3}{2})^2 + \frac{1}{2}\) shows the minimum value is \(\frac{1}{2}\) when \(x = \frac{3}{2}\). The maximum value is \(13\) when \(x = 4\). Therefore, the range is \(\frac{1}{2} \leq g(x) \leq 13\).

(iv) For \(h\) to have an inverse, it must be one-to-one. This occurs when \(x \geq \frac{3}{2}\), so the smallest \(k = \frac{3}{2}\).

(v) For \(k = \frac{3}{2}\), express \(h(x) = 2(x - \frac{3}{2})^2 + \frac{1}{2}\). To find \(h^{-1}(x)\), solve for \(x\):

\(y = 2(x - \frac{3}{2})^2 + \frac{1}{2}\)

\(y - \frac{1}{2} = 2(x - \frac{3}{2})^2\)

\(\frac{y - \frac{1}{2}}{2} = (x - \frac{3}{2})^2\)

\(x - \frac{3}{2} = \pm \sqrt{\frac{y - \frac{1}{2}}{2}}\)

Since \(x \geq \frac{3}{2}\), choose the positive root:

\(x = \frac{3}{2} + \sqrt{\frac{y - \frac{1}{2}}{2}}\)

Thus, \(h^{-1}(x) = \frac{3}{2} + \sqrt{\frac{x}{2} - \frac{1}{4}}\).