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

\(V = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = \frac{4}{2x-1}\):

\(V = \pi \int_{1}^{2} \left( \frac{4}{2x-1} \right)^2 \, dx = \pi \int_{1}^{2} \frac{16}{(2x-1)^2} \, dx\)

Integrate:

\(\int \frac{16}{(2x-1)^2} \, dx = -\frac{16}{2x-1}\)

Evaluate from 1 to 2:

\(V = \pi \left[ -\frac{16}{2x-1} \right]_{1}^{2} = \pi \left( -\frac{16}{3} + 16 \right) = \frac{16\pi}{3}\)

(ii) The slope of the tangent to the curve is given by the derivative:

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

The slope of the normal is \(m = -\frac{1}{2} \times (-2) = 2\).

Equating \(\frac{dy}{dx} = -2\):

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

Solve for \(x\):

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

For \(x = \frac{3}{2}\), \(y = 2\), so \(2y = x + c\) gives \(4 = \frac{3}{2} + c\), \(c = \frac{5}{2}\).

For \(x = -\frac{1}{2}\), \(y = -2\), so \(2y = x + c\) gives \(-4 = -\frac{1}{2} + c\), \(c = -\frac{7}{2}\).