(i) To find the equation of the tangent at A (2, 2), first find the derivative of the curve \(y = x^2 - 2x + 2\), which is \(\frac{dy}{dx} = 2x - 2\).

At \(x = 2\), the gradient \(m\) is \(2(2) - 2 = 2\).

The equation of the tangent is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\).

(ii) The equation of the normal at A is \(y - 2 = -\frac{1}{2}(x - 2)\), which simplifies to \(y = -\frac{1}{2}x + 3\).

Substitute \(y = -\frac{1}{2}x + 3\) into the curve equation \(x^2 - 2x + 2 = y\) to find \(B\):

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

\(x^2 - 2x + \frac{1}{2}x - 1 = 0\)

\(2x^2 - 3x - 2 = 0\)

Solving this quadratic gives \(x = -\frac{1}{2}\), and substituting back gives \(y = \frac{3}{4}\).

Thus, \(B = \left(-\frac{1}{2}, \frac{3}{4}\right)\).

(iii) At \(x = -\frac{1}{2}\), the gradient is \(2(-\frac{1}{2}) - 2 = -3\).

The equation of the tangent at B is \(y - \frac{3}{4} = -3(x + \frac{1}{2})\), which simplifies to \(y = -3x + \frac{7}{4}\).

Equate the tangents \(y = 2x - 2\) and \(y = -3x + \frac{7}{4}\):

\(2x - 2 = -3x + \frac{7}{4}\)

\(2x + 3x = \frac{7}{4} + 2\)

\(5x = \frac{15}{4}\)

\(x = \frac{3}{4}\)

Substitute \(x = \frac{3}{4}\) into \(y = 2x - 2\) to find \(y = -\frac{1}{2}\).

Thus, \(C = \left(\frac{3}{4}, -\frac{1}{2}\right)\).

(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.