(i) To show that \(PQ\) is a normal to the curve, we first find the derivative of the curve \(y = (1 + 4x)^{\frac{1}{2}}\).

The derivative is \(\frac{dy}{dx} = \frac{1}{2}(1 + 4x)^{-\frac{1}{2}} \times 4 = 2(1 + 4x)^{-\frac{1}{2}}\).

At \(x = 6\), \(\frac{dy}{dx} = \frac{2}{5}\).

The gradient of the normal at \(P\) is \(-\frac{1}{2}\).

The gradient of \(PQ\) is \(-\frac{5}{2}\), hence \(PQ\) is a normal, or \(m_1 m_2 = -1\).

(ii) To find the volume of revolution, we calculate the volume for the curve and the line separately.

Volume for the curve: \(V = \pi \int (1 + 4x) \, dx\).

Integrating \(y^2 = (1 + 4x)\), we get \(\pi \left[ x + 2x^2 \right]\) from 0 to 6.

\(= \pi [6 + 72 - 0] = 78\pi\).

Volume for the line: \(V = \frac{1}{3} \pi \times 5^2 \times 2 = \frac{50}{3} \pi\).

Total volume: \(78\pi + \frac{50}{3}\pi = 94\frac{2}{3}\pi\) or \(\frac{284\pi}{3}\).