Start with the function \(y = h(x) = 4x^2 - 12x + 13\).

Complete the square for the quadratic expression:

\(4x^2 - 12x + 13 = 4(x^2 - 3x) + 13\).

\(= 4((x - \frac{3}{2})^2 - \frac{9}{4}) + 13\).

\(= 4(x - \frac{3}{2})^2 - 9 + 13\).

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

Set \(y = 4(x - \frac{3}{2})^2 + 4\).

Rearrange to solve for \(x\):

\(y - 4 = 4(x - \frac{3}{2})^2\).

\((x - \frac{3}{2})^2 = \frac{y - 4}{4}\).

\(x - \frac{3}{2} = \pm \sqrt{\frac{y - 4}{4}}\).

Since \(x < 0\), choose the negative root:

\(x = \frac{3}{2} - \frac{\sqrt{y - 4}}{2}\).

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

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

\(7 - 2(x^2 + 6x) = 7 - 2((x + 3)^2 - 9) = 7 - 2(x + 3)^2 + 18 = 25 - 2(x + 3)^2\).

Thus, \(a = 25\) and \(b = 3\).

(ii) The stationary point occurs where the derivative is zero. The derivative of \(f(x) = 7 - 2x^2 - 12x\) is \(f'(x) = -4x - 12\).

Setting \(f'(x) = 0\), we get \(-4x - 12 = 0\) which gives \(x = -3\).

Substitute \(x = -3\) into \(f(x)\):

\(f(-3) = 7 - 2(-3)^2 - 12(-3) = 7 - 18 + 36 = 25\).

Thus, the stationary point is \((-3, 25)\).

(iii) For \(g\) to have an inverse, it must be one-to-one. The function \(g(x) = 7 - 2x^2 - 12x\) is a downward-opening parabola, so it is one-to-one for \(x \geq -3\).

Thus, the smallest value of \(k\) is \(-3\).

(iv) To find \(g^{-1}(x)\), start with \(y = 7 - 2x^2 - 12x\) and solve for \(x\):

Rearrange to \(2x^2 + 12x + y - 7 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 12\), \(c = y - 7\):

\(x = \frac{-12 \pm \sqrt{12^2 - 8(y - 7)}}{4} = \frac{-12 \pm \sqrt{144 - 8y + 56}}{4} = \frac{-12 \pm \sqrt{200 - 8y}}{4}\).

Since \(x \geq -3\), choose the positive root:

\(x = \frac{-12 + \sqrt{200 - 8y}}{4}\).

Thus, \(g^{-1}(x) = \sqrt{\frac{25 - x}{2}} - 3\).

(iii) To express \(6x - x^2 - 5\) in the form \(a - (x - b)^2\), we complete the square:

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

\(= -(x^2 - 6x + 9 - 9) - 5\)

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

\(= 4 - (x - 3)^2\)

Thus, \(a = 4\) and \(b = 3\).

(iv) The function \(g\) has an inverse when it is one-to-one. The vertex of the parabola \(y = 6x - x^2 - 5\) is at \(x = 3\). Therefore, the smallest value of \(k\) for which \(g\) is one-to-one is \(k = 3\).

(v) For \(k = 3\), the function \(g(x) = 4 - (x - 3)^2\). To find the inverse, set \(y = 4 - (x - 3)^2\) and solve for \(x\):

\(y = 4 - (x - 3)^2\)

\((x - 3)^2 = 4 - y\)

\(x - 3 = \pm \sqrt{4 - y}\)

Since \(x \geq 3\), we take the positive root:

\(x = \sqrt{4 - y} + 3\)

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

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

To find the smallest possible value of \(k\), we need the function \(h(x) = x^2 + 4x\) to be one-to-one, which occurs when it is either strictly increasing or decreasing. The vertex of the parabola \(x^2 + 4x\) is at \(x = -\frac{b}{2a} = -\frac{4}{2} = -2\). Thus, the smallest possible value of \(k\) is \(-2\) so that the function is increasing for \(x \geq -2\).

To find \(h^{-1}(x)\), start by setting \(y = x^2 + 4x\).

Complete the square: \(y = (x+2)^2 - 4\).

Rearrange to solve for \(x\):

\(y + 4 = (x+2)^2\)

\(\sqrt{y + 4} = x + 2\)

\(x = \sqrt{y + 4} - 2\)

Thus, the inverse function is \(h^{-1}(x) = \sqrt{x+4} - 2\).