(i) To express \(8x - x^2\) in the form \(a - (x + b)^2\), we complete the square:

\(8x - x^2 = -(x^2 - 8x) = -(x^2 - 8x + 16 - 16) = -(x - 4)^2 + 16\).

Thus, \(a = 16\) and \(b = -4\).

(ii) The stationary point occurs where \(\frac{dy}{dx} = 0\). Differentiating, \(\frac{dy}{dx} = 8 - 2x\).

Setting \(8 - 2x = 0\) gives \(x = 4\).

Substituting \(x = 4\) into \(y = 8x - x^2\), we get \(y = 16\).

Thus, the stationary point is \((4, 16)\).

(iii) Solve \(8x - x^2 \geq -20\):

\(x^2 - 8x - 20 \leq 0\).

Factorizing gives \((x - 10)(x + 2) \leq 0\).

The solution is \(-2 \leq x \leq 10\).

(iv) From the function \(g(x) = 8x - x^2\) for \(x \geq 4\), the maximum value is at \(x = 4\), giving \(y = 16\).

Thus, the domain of \(g^{-1}\) is \(x \leq 16\) and the range is \(x \geq 4\).

(v) To find \(g^{-1}(x)\), solve \(y = 8x - x^2\) for \(x\):

\(x^2 - 8x + y = 0\).

Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 4y}}{2}\).

Since \(x \geq 4\), we take the positive root: \(x = 4 + \sqrt{16 - y}\).

Thus, \(g^{-1}(x) = 4 + \sqrt{16 - x}\).