(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:

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

(i) Differentiate \(y = (x-1)^{-2} + 2\) to find the gradient of the tangent:

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

At \(x = 2\), \(\frac{dy}{dx} = -2\), so the gradient of the normal is \(\frac{1}{2}\).

The equation of the normal through \((2, 3)\) is:

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

Simplifying gives \(y = \frac{1}{2}x + 2\).

(ii) The volume of revolution is given by:

\(\pi \int_{1}^{3} y_1^2 \, dx + \pi \int_{1}^{3} y_2^2 \, dx\).

For \(y_1 = \frac{1}{2}x + 2\), \(y_1^2 = \left(\frac{1}{2}x + 2\right)^2\).

For \(y_2 = (x-1)^{-2} + 2\), \(y_2^2 = \left((x-1)^{-2} + 2\right)^2\).

Integrate and evaluate:

\(\pi \left[ \frac{1}{3} \left(\frac{1}{2}x + 2\right)^3 \right]_{1}^{3} + \pi \left[ \frac{1}{12}x^2 + x + 4x \right]_{1}^{3}\).

Calculate the definite integrals and sum the volumes:

\(\frac{7}{12} \pi + \frac{6}{7} \pi = \frac{111}{8} \pi\) or \(13.9\pi\) or 43.6.

(i) To express \(y = x^2 + 4x + 3\) in the form \(y = (x + a)^2 + b\), complete the square:

\(y = (x + 2)^2 - 1\).

Hence, \(a = 2\) and \(b = -1\).

To express \(x\) in terms of \(y\):

\(y = (x + 2)^2 - 1\)

\(y + 1 = (x + 2)^2\)

\(x + 2 = \pm \sqrt{y + 1}\)

Since \(x \geq -2\), we take the positive root:

\(x = -2 + \sqrt{y + 1}\).

(ii) To find the volume when the shaded region is rotated about the y-axis, use the formula for the volume of revolution:

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

Substitute \(x = -2 + \sqrt{y + 1}\):

\(x^2 = 4 + (y + 1) - 4(y + 1)^{1/2}\)

\(V = \pi \int_{-1}^{3} \left[ 5y + \frac{y^2}{2} - \frac{4(y + 1)^{3/2}}{3/2} \right] \, dy\)

Integrate and apply limits:

\(V = \pi \left[ 15 + \frac{9}{2} - \frac{64}{3} - \left(-5 + \frac{1}{2}\right) \right]\)

\(V = \frac{8\pi}{3}\) or 8.38

(i) To find the volume of the solid obtained by rotating the region about the x-axis, we use the formula:

\(V = \pi \int_0^3 (x^3 + x^2) \, dx\)

Integrating, we have:

\(\pi \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_0^3\)

\(= \pi \left[ \frac{81}{4} + 9 - 0 \right]\)

\(= \frac{117\pi}{4}\)

(ii) First, find the derivative \(\frac{dy}{dx}\):

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

At \(x = 3\), \(y = 6\).

The gradient \(m\) at \(x = 3\) is:

\(m = \frac{1}{2} \times \frac{1}{6} \times 33 = \frac{11}{4}\)

The equation of the normal is:

\(y - 6 = -\frac{4}{11}(x - 3)\)

When \(x = 0\), \(y =7 \frac{1}{11}\).

(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:

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

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