(i) To find \(\frac{dy}{dx}\), use the chain rule. Let \(u = 8x - x^2\), then \(y = \sqrt{u} = u^{\frac{1}{2}}\).

\(\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}\)

\(\frac{du}{dx} = 8 - 2x\)

Thus, \(\frac{dy}{dx} = \frac{1}{2}(8x-x^2)^{-\frac{1}{2}} \times (8-2x)\).

For the stationary point, set \(\frac{dy}{dx} = 0\):

\((8-2x) = 0 \Rightarrow x = 4\).

Substitute \(x = 4\) into \(y = \sqrt{(8x - x^2)}\):

\(y = \sqrt{(8 \times 4 - 4^2)} = \sqrt{16} = 4\).

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

(ii) To find the volume, use the formula for the volume of revolution:

\(V = \pi \int (8x - x^2) \, dx\) from \(x = 0\) to \(x = 8\).

\(V = \pi \left[ \frac{4x^2}{2} - \frac{x^3}{3} \right]_0^8\)

\(V = \pi \left[ \frac{4 \times 8^2}{2} - \frac{8^3}{3} \right]\)

\(V = \pi \left[ 128 - \frac{512}{3} \right]\)

\(V = \pi \left[ \frac{384}{3} \right]\)

\(V = \frac{256\pi}{3}\).