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