(i) To find the volume of the solid of revolution, use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Substitute \(y = 2(3x - 1)^{-\frac{1}{3}}\):

\(V = \pi \int_{\frac{2}{3}}^{3} \left(2(3x - 1)^{-\frac{1}{3}}\right)^2 \, dx\)

\(V = 4\pi \int_{\frac{2}{3}}^{3} (3x - 1)^{-\frac{2}{3}} \, dx\)

Integrate:

\(V = 4\pi \left[ \frac{(3x - 1)^{\frac{1}{3}}}{\frac{1}{3}} \right]_{\frac{2}{3}}^{3}\)

\(V = 4\pi \left[ 3(3x - 1)^{\frac{1}{3}} \right]_{\frac{2}{3}}^{3}\)

Apply the limits:

\(V = 4\pi [2 - 1]\)

\(V = 4\pi\)

Thus, the volume is \(4\pi\) or approximately 12.6.

(ii) Differentiate \(y = 2(3x - 1)^{-\frac{1}{3}}\) to find the gradient:

\(\frac{dy}{dx} = (-\frac{2}{3})(3x - 1)^{-\frac{4}{3}} \times 3\)

\(\frac{dy}{dx} = -2(3x - 1)^{-\frac{4}{3}}\)

At \(x = \frac{2}{3}\), \(y = 2\), and \(\frac{dy}{dx} = -2\).

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

The equation of the normal is:

\(y - 2 = \frac{1}{2}(x - \frac{2}{3})\)

\(y = \frac{1}{2}x + \frac{5}{3}\)