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

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

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

\(V = \pi \int_{1}^{\frac{3}{2}} \left( \frac{4}{(2x-1)^2} \right)^2 \, dx = \pi \int_{1}^{\frac{3}{2}} \frac{16}{(2x-1)^4} \, dx\)

Integrate:

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

Calculate the definite integral:

\(\pi \left( \frac{112}{48} \right) = \frac{11}{6} \pi\)

(b) Differentiate \(y = \frac{4}{(2x-1)^2}\) to find the gradient at \(B\):

\(\frac{dy}{dx} = -8(2x-1)^{-3} \times 2\)

At \(B\), the gradient is \(-2\).

Equation of the tangent at \(B\):

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

Equation of the normal at \(B\):

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

The tangent crosses the x-axis at 2, and the normal crosses the x-axis at \(\frac{1}{2}\).

Calculate the area of the triangle:

\(\text{Area} = \frac{1}{2} \times \left(2 - \frac{1}{2}\right) \times 1 = \frac{5}{4}\)