(i) To find the stationary point, set \(f'(x) = 0\):

\(ax^2 + bx = 0\)

\(x(ax + b) = 0\)

\(x = 0\) or \(x = -\frac{b}{a}\)

The non-zero value is \(x = -\frac{b}{a}\).

To determine the nature, find \(f''(x)\):

\(f''(x) = 2ax + b\)

Substitute \(x = -\frac{b}{a}\) into \(f''(x)\):

\(f''\left(-\frac{b}{a}\right) = 2a\left(-\frac{b}{a}\right) + b = -2b + b = -b\)

Since \(f''\left(-\frac{b}{a}\right) < 0\), it is a maximum point.

(ii) Given \(f'(-2) = 0\) and \(f'(1) = 9\):

\(a(-2)^2 + b(-2) = 0\)

\(4a - 2b = 0\)

\(2a = b\)

\(a(1)^2 + b(1) = 9\)

\(a + b = 9\)

Substitute \(b = 2a\) into \(a + b = 9\):

\(a + 2a = 9\)

\(3a = 9\)

\(a = 3\)

\(b = 6\)

Integrate \(f'(x) = 3x^2 + 6x\) to find \(f(x)\):

\(f(x) = \int (3x^2 + 6x) \, dx = x^3 + 3x^2 + c\)

Given \(f(-2) = -3\):

\(-3 = (-2)^3 + 3(-2)^2 + c\)

\(-3 = -8 + 12 + c\)

\(c = -7\)

Thus, \(f(x) = x^3 + 3x^2 - 7\).

(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 \(-3.83\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 7 - x^2 - 6x\):

\(y = \int (7 - x^2 - 6x) \, dx = 7x - \frac{x^3}{3} - \frac{6x^2}{2} + c\).

Using the point \((3, -10)\), substitute to find \(c\):

\(-10 = 7(3) - \frac{3^3}{3} - \frac{6(3)^2}{2} + c\)

\(-10 = 21 - 9 - 27 + c\)

\(c = 5\)

Thus, the equation is \(y = 7x - \frac{x^3}{3} - \frac{6x^2}{2} + 5\).

(ii) Express \(7 - x^2 - 6x\) in the form \(a - (x + b)^2\):

Complete the square for \(-x^2 - 6x\):

\(-x^2 - 6x = -(x^2 + 6x) = -(x + 3)^2 + 9\)

Thus, \(7 - x^2 - 6x = 7 + 9 - (x + 3)^2 = 16 - (x + 3)^2\)

So, \(a = 16\) and \(b = 3\).

(iii) Find the set of values of \(x\) for which the gradient is positive:

\(7 - x^2 - 6x > 0\)

Using the completed square form: \(16 - (x + 3)^2 > 0\)

\((x + 3)^2 < 16\)

\(-4 < x + 3 < 4\)

\(-7 < x < 1\)

(i) At \(x = a^2\), \(\frac{dy}{dx} = \frac{2}{a}a^{-1} + a \cdot a^{-3} = \frac{3}{a^2}\). The equation of the tangent is \(y - 3 = \frac{3}{a^2}(x - a^2)\), simplifying to \(y = \frac{3}{a^2}x\).

(ii) Integrate \(\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-\frac{3}{2}}\) to get \(y = \frac{2}{a} \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + a \cdot \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + c = \frac{4x^{\frac{1}{2}}}{a} - 2ax^{-\frac{1}{2}} + c\). Substituting \(x = a^2, y = 3\) gives \(c = 1\), so \(y = \frac{4x^{\frac{1}{2}}}{a} - 2ax^{-\frac{1}{2}} + 1\).

(iii) Substituting \(x = 16, y = 8\) into the curve equation gives \(8 = \frac{4}{a} \times 4 - 2a \times \frac{1}{4} + 1\). Solving \(a^2 + 14a - 32 = 0\) gives \(a = 2\). Points are \(A = (4, 3)\), \(B = (16, 8)\). Distance \(AB = \sqrt{(16-4)^2 + (8-3)^2} = 13\).

(i) To find the \(x\)-coordinate of \(A\), set \(f'(x) = -1\):

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

Multiply through by \(x^{\frac{1}{2}}\) to clear the fraction:

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

Rearrange to form a quadratic equation:

\(3x + 1 = \frac{2}{x}\)

\(3x^2 + x - 2 = 0\)

Let \(z = x^{\frac{1}{2}}\), then \(z^2 = x\):

\(3z^2 + z - 2 = 0\)

Solving this quadratic gives \(z = \frac{2}{3}\) or \(z = -1\).

Since \(z = x^{\frac{1}{2}}\), \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

(ii) To find the \(y\)-coordinate of \(A\), integrate \(f'(x)\) to find \(f(x)\):

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

\(f(x) = 3 \cdot \frac{2}{3} x^{\frac{3}{2}} - 2 \cdot 2 x^{\frac{1}{2}} + c\)

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

Given \(f(4) = 10\):

\(10 = 2(4)^{\frac{3}{2}} - 4(4)^{\frac{1}{2}} + c\)

\(10 = 16 - 8 + c\)

\(c = 2\)

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

Substitute \(x = \frac{4}{9}\) to find \(y\):

\(y = 2 \left(\frac{4}{9}\right)^{\frac{3}{2}} - 4 \left(\frac{4}{9}\right)^{\frac{1}{2}} + 2\)

\(y = 2 \cdot \frac{8}{27} - 4 \cdot \frac{2}{3} + 2\)

\(y = \frac{16}{27} - \frac{8}{3} + 2\)

\(y = \frac{16}{27} - \frac{72}{27} + \frac{54}{27}\)

\(y = -\frac{2}{27}\)