(i) To find \(f'(x)\), use the chain rule. Let \(u = 4x + 1\), then \(f(x) = u^{\frac{3}{2}}\).

\(\frac{du}{dx} = 4\)

\(\frac{df}{du} = \frac{3}{2}u^{\frac{1}{2}}\)

Thus, \(f'(x) = \frac{3}{2}(4x + 1)^{\frac{1}{2}} \times 4 = 6(4x + 1)^{\frac{1}{2}}\).

To find \(f''(x)\), differentiate \(f'(x)\):

\(f'(x) = 6(4x + 1)^{\frac{1}{2}}\)

\(f''(x) = 6 \times \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \times 4 = 12(4x + 1)^{-\frac{1}{2}}\).

(ii) Calculate \(f(2)\), \(f'(2)\), and \(f''(2)\):

\(f(2) = (4(2) + 1)^{\frac{3}{2}} = 9^{\frac{3}{2}} = 27\)

\(f'(2) = 6(4(2) + 1)^{\frac{1}{2}} = 6 \times 3 = 18\)

\(f''(2) = 12(4(2) + 1)^{-\frac{1}{2}} = 12 \times \frac{1}{3} = 4\)

The terms form a geometric progression: \(27, 18, 4k\).

The common ratio \(r\) is \(\frac{18}{27} = \frac{2}{3}\).

Thus, \(\frac{4k}{18} = \frac{2}{3}\).

Solving for \(k\):

\(4k = 12\)

\(k = 3\)