(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{4x^2 + 12}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}\).

Multiply through by \((x+1)(x-3)^2\) to clear the denominators:

\(4x^2 + 12 = A(x-3)^2 + B(x+1)(x-3) + C(x+1)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = 1, B = 3, C = 12\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fraction decomposition:

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

\(\frac{1}{x-3} = -\frac{1}{3} - \frac{1}{9}x - \ldots\)

\(\frac{1}{(x-3)^2} = \frac{1}{9} + \frac{2}{27}x + \ldots\)

Combine these expansions:

\(f(x) = \frac{1}{x+1} + 3\left(\frac{1}{x-3}\right) + 12\left(\frac{1}{(x-3)^2}\right)\)

\(= \left(1 - x + x^2\right) + 3\left(-\frac{1}{3} - \frac{1}{9}x\right) + 12\left(\frac{1}{9} + \frac{2}{27}x\right)\)

\(= \frac{4}{3} - \frac{4}{9}x + \frac{4}{3}x^2\).