(i) To find the stationary points, set \(\frac{dy}{dx} = 0\):

\(-x^2 + 5x - 4 = 0\)

Solving the quadratic equation:

\(x^2 - 5x + 4 = 0\)

\((x - 1)(x - 4) = 0\)

Thus, \(x = 1\) and \(x = 4\).

(ii) Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -2x + 5\)

Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 1\):

\(\frac{d^2y}{dx^2} = -2(1) + 5 = 3\)

Since \(3 > 0\), \(x = 1\) is a minimum.

Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 4\):

\(\frac{d^2y}{dx^2} = -2(4) + 5 = -3\)

Since \(-3 < 0\), \(x = 4\) is a maximum.

To find the minimum point of the curve \(y = \frac{9}{4}x^2 - 12x + 18\), we can complete the square or use calculus.

First, we find the derivative \(\frac{dy}{dx} = \frac{9}{2}x - 12\).

Setting \(\frac{dy}{dx} = 0\) gives \(\frac{9}{2}x - 12 = 0\).

Solving for \(x\), we get \(x = \frac{24}{9} = \frac{8}{3}\).

Substitute \(x = \frac{8}{3}\) back into the original equation:

\(y = \frac{9}{4} \left( \frac{8}{3} \right)^2 - 12 \left( \frac{8}{3} \right) + 18\).

Calculate \(y = \frac{9}{4} \times \frac{64}{9} - 32 + 18\).

\(y = 16 - 32 + 18 = 2\).

Thus, the coordinates of the minimum point are \(\left( \frac{8}{3}, 2 \right)\).

(i) Differentiate \(y = 8\sqrt{x} - 2x\) to find \(\frac{dy}{dx} = 4x^{-\frac{1}{2}} - 2\). Set \(\frac{dy}{dx} = 0\) to find the stationary point: \(4x^{-\frac{1}{2}} - 2 = 0\). Solving gives \(\sqrt{x} = 2\), so \(x = 4\). Substitute back to find \(y = 8 \sqrt{4} - 2 \times 4 = 8\). The stationary point is \((4, 8)\).

(ii) Differentiate again to find \(\frac{d^2y}{dx^2} = -2x^{-\frac{3}{2}}\). At \(x = 4\), \(\frac{d^2y}{dx^2} = -\frac{1}{4}\), which is negative, indicating a maximum.

(iii) Set \(y = 6\) in the curve equation: \(8\sqrt{x} - 2x = 6\). Rearrange to form a quadratic in \(\sqrt{x}\): \(2u^2 - 8u + 6 = 0\) where \(u = \sqrt{x}\). Solve to find \(u = 1\) or \(u = 3\), giving \(x = 1\) or \(x = 9\).

(iv) For the line \(y = k\) not to meet the curve, \(k > 8\) since the maximum value of \(y\) on the curve is 8.

(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 \(k\):

\(4k = 12\)

\(k = 3\)

To find the stationary points, we first find the derivative \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = 8 + [-2] [(2x-1)^{-2}]\).

Set \(\frac{dy}{dx} = 0\):

\(0 = 8 - \frac{2}{(2x-1)^2}\).

Rearrange to find \((2x-1)^2 = \frac{1}{4}\).

Solving gives \(x = \frac{1}{4}\) and \(x = \frac{3}{4}\).

Next, find the second derivative \(\frac{d^2y}{dx^2} = 8(2x-1)^{-3}\).

Evaluate \(\frac{d^2y}{dx^2}\) at each stationary point:

When \(x = \frac{1}{4}\), \(\frac{d^2y}{dx^2} = -64\), which is less than 0, indicating a maximum.

When \(x = \frac{3}{4}\), \(\frac{d^2y}{dx^2} = 64\), which is greater than 0, indicating a minimum.