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

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

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

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

Expand and collect terms:

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

Equate coefficients to solve for \(A\) and \(B\):

\(3A + 2B = 12\) and \(-2A + B = 12\).

Solving these equations gives \(A = 2\), \(B = -4\), and \(C = 6\).

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

(ii) Expansion in ascending powers of \(x\):

Expand \(\frac{1}{2+x}\) using the binomial series:

\(\frac{1}{2+x} = \frac{1}{2}(1 - \frac{x}{2} + (\frac{x}{2})^2 + \ldots) = \frac{1}{2} - \frac{x}{4} + \frac{x^2}{8} + \ldots\).

Expand \(\frac{1}{3-2x}\) using the binomial series:

\(\frac{1}{3-2x} = \frac{1}{3}(1 + \frac{2x}{3} + (\frac{2x}{3})^2 + \ldots) = \frac{1}{3} + \frac{2x}{9} + \frac{4x^2}{27} + \ldots\).

Substitute back into the partial fractions:

\(f(x) = 2(\frac{1}{2} - \frac{x}{4} + \frac{x^2}{8}) - 4(\frac{1}{3} + \frac{2x}{9} + \frac{4x^2}{27})\).

Simplify to obtain:

\(f(x) = 2 + \frac{7}{3}x + \frac{7}{18}x^2\).