(a) To find \(k\), first find the derivative \(f'(x) = 2x - \frac{k}{x^2}\).

Set \(f'(2) = 0\):

\(2 \times 2 - \frac{k}{2^2} = 0\)

\(4 - \frac{k}{4} = 0\)

\(k = 16\)

(b) To determine the nature of the stationary point, find the second derivative \(f''(x) = 2 + \frac{2k}{x^3}\).

Evaluate \(f''(2)\):

\(f''(2) = 2 + \frac{2 \times 16}{2^3} = 2 + \frac{32}{8} = 6\)

Since \(f''(2) > 0\), the stationary point is a minimum.

(c) When \(x = 2\), \(f(x) = 2^2 + \frac{16}{2} + 2 = 4 + 8 + 2 = 14\).

Since this is the only stationary point and it is a minimum, the range of \(f\) is \(y \geq 14\).