(i) To find \(f'(x)\), integrate \(f''(x) = (4x + 1)^{-\frac{1}{2}}\):

\(f'(x) = \int (4x + 1)^{-\frac{1}{2}} \, dx = \frac{1}{2}(4x + 1)^{\frac{1}{2}} + c\).

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

\(\frac{1}{2}(4(2) + 1)^{\frac{1}{2}} + c = 0 \Rightarrow \frac{3}{2} + c = 0 \Rightarrow c = -\frac{3}{2}\).

Thus, \(f'(x) = \frac{1}{2}(4x + 1)^{\frac{1}{2}} - \frac{3}{2}\).

(ii) Given \(f''(0), f'(0), f(0)\) form an arithmetic progression:

\(f''(0) = 1\), \(f'(0) = \frac{1}{2} - \frac{3}{2} = -1\).

Arithmetic progression: \(1, -1, f(0)\).

Common difference \(d = -2\), so \(f(0) = -1 - 2 = -3\).

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

\(f(x) = \int \left( \frac{1}{2}(4x + 1)^{\frac{1}{2}} - \frac{3}{2} \right) \, dx\).

\(f(x) = \frac{1}{12}(4x + 1)^{\frac{3}{2}} - \frac{3}{2}x + k\).

Given \(f(0) = -3\), substitute \(x = 0\):

\(\frac{1}{12}(1)^{\frac{3}{2}} - 0 + k = -3 \Rightarrow \frac{1}{12} + k = -3 \Rightarrow k = -\frac{37}{12}\).

Thus, \(f(x) = \frac{1}{12}(4x + 1)^{\frac{3}{2}} - \frac{3}{2}x - \frac{37}{12}\).

Minimum value at \(x = 2\):

\(f(2) = \frac{1}{12}(9)^{\frac{3}{2}} - 3 - \frac{37}{12} = \frac{27}{12} - 3 - \frac{37}{12} = \frac{23}{6}\) or \(-3.83\).