(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.