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

Assume \(\frac{5x^2 + x + 6}{(3 - 2x)(x^2 + 4)} = \frac{A}{3 - 2x} + \frac{Bx + C}{x^2 + 4}\).

Multiply through by the denominator \((3 - 2x)(x^2 + 4)\) to get:

\(5x^2 + x + 6 = A(x^2 + 4) + (Bx + C)(3 - 2x)\).

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

\(A = 3, B = -1, C = -2\).

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

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

For \(\frac{3}{3 - 2x}\), use the expansion \((3 - 2x)^{-1} = \frac{1}{3}(1 - \frac{2}{3}x)^{-1}\).

First two terms: \(\frac{1}{3}(1 + \frac{2}{3}x + (\frac{2}{3}x)^2) = \frac{1}{3} + \frac{2}{9}x + \frac{4}{27}x^2\).

For \(\frac{-x - 2}{x^2 + 4}\), use the expansion \((x^2 + 4)^{-1} = \frac{1}{4}(1 - \frac{1}{4}x^2)\).

First two terms: \(\frac{-x - 2}{4}(1 - \frac{1}{4}x^2) = \frac{-x}{4} - \frac{1}{2} + \frac{x^3}{16}\).

Combine terms up to \(x^2\):

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

Simplify to get the final answer:

\(\frac{1}{2} + \frac{5}{12}x + \frac{41}{72}x^2\).