(i) To determine whether the stationary point is a maximum or minimum, evaluate the second derivative at \(x = 3\):

\(f''(3) = 36 \times 3^{-3} = \frac{36}{27} = \frac{4}{3}\).

Since \(f''(3) > 0\), the stationary point at \((3, 7)\) is a minimum.

(ii) To find \(f'(x)\), integrate \(f''(x)\):

\(f''(x) = 36x^{-3}\) implies \(f'(x) = \int 36x^{-3} \, dx = -18x^{-2} + c\).

Given that \(f'(3) = 0\), substitute to find \(c\):

\(0 = -18 \times 3^{-2} + c\)

\(0 = -2 + c\)

\(c = 2\)

Thus, \(f'(x) = -18x^{-2} + 2\).

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

\(f(x) = \int (-18x^{-2} + 2) \, dx = 18x^{-1} + 2x + k\).

Given that \(f(3) = 7\), substitute to find \(k\):

\(7 = 18 \times 3^{-1} + 2 \times 3 + k\)

\(7 = 6 + 6 + k\)

\(k = -5\)

Thus, \(f(x) = 18x^{-1} + 2x - 5\).

(i) At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{24}{2^3} - 4 = 3 - 4 = -1\), which is negative, indicating a maximum point.

(ii) To find \(\frac{dy}{dx}\), integrate \(\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4\):

\(\frac{dy}{dx} = \int \left( \frac{24}{x^3} - 4 \right) dx = \int 24x^{-3} dx - \int 4 dx\)

\(= -12x^{-2} - 4x + A\)

Given \(\frac{dy}{dx} = 0\) at \(x = 2\), solve for \(A\):

\(0 = -12(2)^{-2} - 4(2) + A\)

\(0 = -3 - 8 + A\)

\(A = 11\)

Thus, \(\frac{dy}{dx} = -12x^{-2} - 4x + 11\).

(iii) Integrate \(\frac{dy}{dx} = -12x^{-2} - 4x + 11\) to find \(y\):

\(y = \int (-12x^{-2} - 4x + 11) dx\)

\(= 12x^{-1} - 2x^2 + 11x + c\)

Given \(y = 13\) when \(x = 1\):

\(13 = 12(1)^{-1} - 2(1)^2 + 11(1) + c\)

\(13 = 12 - 2 + 11 + c\)

\(c = -8\)

Thus, \(y = 12x^{-1} - 2x^2 + 11x - 8\).

At \(x = 2\), \(y = 12(2)^{-1} - 2(2)^2 + 11(2) - 8\)

\(y = 6 - 8 + 22 - 8 = 12\)

Therefore, the coordinates of the stationary point \(P\) are \((2, 12)\).

(i) Calculate f'(2):

\(f'(2) = 2(2) - \frac{2}{2^2} = 4 - \frac{1}{2} = \frac{7}{2}\)

The gradient of the normal is the negative reciprocal:

\(-\frac{2}{7}\)

Equation of the normal using point-slope form:

\(y - 6 = -\frac{2}{7}(x - 2)\)

(ii) Integrate f'(x) to find f(x):

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

Substitute point (2, 6):

\(6 = 4 + 1 + c \Rightarrow c = 1\)

Thus, \(f(x) = x^2 + \frac{2}{x} + 1\)

\((iii) Find stationary points by setting f'(x) = 0:\)

\(2x - \frac{2}{x^2} = 0 \Rightarrow 2x^3 - 2 = 0 \Rightarrow x = 1\)

Determine nature using second derivative:

\(f''(x) = 2 + \frac{4}{x^3}\)

\(f''(1) = 6\) (which is > 0), hence minimum.

(i) The slope of the normal is \(-\frac{1}{3}\), so the slope of the tangent is \(3\) (since \(m_1 m_2 = -1\)).

Set \(\frac{dy}{dx} = 3 = \frac{12}{\sqrt{4 \times 2 + a}}\).

\(3 = \frac{12}{\sqrt{8 + a}}\)

\(\sqrt{8 + a} = 4\)

\(8 + a = 16\)

\(a = 8\)

(ii) Integrate \(\frac{dy}{dx} = \frac{12}{\sqrt{4x + 8}}\) to find \(y\).

\(y = \int 12(4x + 8)^{-\frac{1}{2}} \, dx\)

\(y = 12 \times 2 \sqrt{4x + 8} + c\)

\(y = 24 \sqrt{4x + 8} + c\)

Using point \(P(2, 14)\):

\(14 = 24 \sqrt{16} + c\)

\(14 = 24 \times 4 + c\)

\(14 = 96 + c\)

\(c = -82\)

Thus, the equation of the curve is \(y = 24 \sqrt{4x + 8} - 82\).

We start with the second derivative \(\frac{d^2y}{dx^2} = 2x - 1\). To find the first derivative \(\frac{dy}{dx}\), integrate:

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

Given the minimum point at \(x = 3\), \(\frac{dy}{dx} = 0\) at this point:

\(0 = 3^2 - 3 + c\) which simplifies to \(c = -6\).

Thus, \(\frac{dy}{dx} = x^2 - x - 6\).

Setting \(\frac{dy}{dx} = 0\) to find critical points:

\(x^2 - x - 6 = 0\).

Factoring gives \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\).

Since \(x = 3\) is a minimum, \(x = -2\) is a maximum.

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (x^2 - x - 6) \, dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + k\).

Given \(y = -10\) when \(x = 3\), substitute to find \(k\):

\(-10 = \frac{1}{3}(3)^3 - \frac{1}{2}(3)^2 - 6(3) + k\).

\(-10 = 9 - 4.5 - 18 + k\).

\(k = \frac{3}{2}\).

Thus, \(y = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + \frac{3}{2}\).

Substitute \(x = -2\) to find \(y\) at the maximum point:

\(y = \frac{1}{3}(-2)^3 - \frac{1}{2}(-2)^2 - 6(-2) + \frac{3}{2}\).

\(y = -\frac{8}{3} - 2 + 12 + \frac{3}{2}\).

\(y = 10.8\).

Therefore, the coordinates of the maximum point are \((-2, 10.8)\).