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