(i) To find the gradient of the curve at \(x = 2\), we first differentiate \(y = \frac{6}{3x - 2}\).

Using the chain rule, \(\frac{dy}{dx} = -6(3x - 2)^{-2} \times 3\).

Substitute \(x = 2\) into the derivative:

\(\frac{dy}{dx} = -6(3 \times 2 - 2)^{-2} \times 3 = -6(4)^{-2} \times 3 = -\frac{9}{8}\).

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

\(V = \pi \int y^2 \, dx\).

\(V = \pi \int_{1}^{2} \left( \frac{6}{3x - 2} \right)^2 \, dx\).

\(V = \pi \int_{1}^{2} \frac{36}{(3x - 2)^2} \, dx\).

The integral of \(\frac{36}{(3x - 2)^2}\) is \(\frac{-36}{3x - 2}\).

Evaluate from 1 to 2:

\(\left[ \frac{-36}{3x - 2} \right]_{1}^{2} = \left( \frac{-36}{4} - \frac{-36}{1} \right) = -9 + 36 = 27\).

Thus, the volume is \(9\pi\).