(i) Differentiate the curve: \(\frac{dy}{dx} = -(x-1)^{-2} + 9(x-5)^{-2}\).

At \(x = 3\), \(\frac{dy}{dx} = \frac{1}{4} + \frac{9}{4} = 2\).

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

Equation of the normal: \(y - 5 = -\frac{1}{2}(x - 3)\).

Setting \(y = 0\) to find \(x\)-intercept: \(0 - 5 = -\frac{1}{2}(x - 3)\).

Solving gives \(x = 13\).

(ii) Set \(\frac{dy}{dx} = 0\): \((x-5)^2 = 9(x-1)^2\).

Solving gives \(x - 5 = \pm 3(x-1)\) or \((x^2 - x - 2) = 0\).

Solutions are \(x = -1\) and \(x = 2\).

Second derivative: \(\frac{d^2y}{dx^2} = 2(x-1)^{-3} - 18(x-5)^{-3}\).

At \(x = -1\), \(\frac{d^2y}{dx^2} = -\frac{1}{6} < 0\), so maximum.

At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{8}{3} > 0\), so minimum.