(a) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = -\left( \frac{1}{2}x + k \right)^{-3} \cdot \frac{1}{2}\).

Since the curve has a minimum at \(x = 2\), \(f''(2) > 0\):

\(-\left( 1 + k \right)^{-3} > 0\).

This implies \(k < -1\).

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

\(f(x) = \int \left( \left( \frac{1}{2}x - 3 \right)^{-2} - (-2)^{-2} \right) \, dx\).

Using integration,

\(f(x) = \left[ \left( \frac{1}{2}x - 3 \right)^{-1} \cdot \frac{-1}{\frac{1}{2}} \right] - \left[ (-2)^{-1} \cdot x \right] + c\).

Substitute \(x = 2, y = 3\frac{1}{2}\) to find \(c\):

\(3\frac{1}{2} = 1 - \frac{1}{2} + c\).

\(c = 3\).

Thus, \(f(x) = \frac{-2}{\left( \frac{1}{2}x - 3 \right)} \cdot \frac{x}{4} + 3\).

(c) Set \(f'(x) = 0\) to find the other stationary point:

\(\left( \frac{1}{2}x - 3 \right)^{-2} - (-2)^{-2} = 0\).

\(\left( \frac{1}{2}x - 3 \right)^{-2} = 4\).

\(\frac{1}{2}x - 3 = \pm 2\).

Solving gives \(x = 10\).

Substitute \(x = 10\) into \(f(x)\):

\(y = -1 - 2\frac{1}{2} + 3 = -\frac{1}{2}\).

Check the nature using \(f''(10)\):

\(f''(10) = -\left( 5 - 3 \right)^{-3} < 0\), indicating a maximum.