(i) Given \(f''(x) = \frac{12}{x^3}\), integrate to find \(f'(x)\):

\(f'(x) = \int \frac{12}{x^3} \, dx = -\frac{6}{x^2} + c\).

Since \(f'(2) = 0\), substitute \(x = 2\):

\(0 = -\frac{6}{4} + c \Rightarrow c = \frac{3}{2}\).

Thus, \(f'(x) = -\frac{6}{x^2} + \frac{3}{2}\).

Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int \left(-\frac{6}{x^2} + \frac{3}{2}\right) \, dx = \frac{6}{x} + \frac{3x}{2} + A\).

Since \(f(2) = 10\), substitute \(x = 2\):

\(10 = \frac{6}{2} + 3 + A \Rightarrow A = 4\).

Thus, \(f(x) = \frac{6}{x} + \frac{3x}{2} + 4\).

(ii) Set \(f'(x) = 0\):

\(-\frac{6}{x^2} + \frac{3}{2} = 0 \Rightarrow x = \pm 2\).

The other stationary point is \((-2, -2)\).

(iii) Evaluate \(f''(x)\) at the stationary points:

At \(x = 2\), \(f''(2) = \frac{12}{8} = 1.5\), indicating a minimum.

At \(x = -2\), \(f''(-2) = -1.5\), indicating a maximum.