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

Assume \(\frac{2x^2 - 7x - 1}{(x-2)(x^2+3)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+3}\).

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

\(2x^2 - 7x - 1 = A(x^2+3) + (Bx+C)(x-2)\).

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

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

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

Use the partial fractions: \(\frac{-1}{x-2} + \frac{3x-1}{x^2+3}\).

Expand \(\left(1 - \frac{x}{2}\right)^{-1}\) and \((x^2+3)^{-1}\) using binomial series:

\(\left(1 - \frac{x}{2}\right)^{-1} = 1 + \frac{x}{2} + \cdots\)

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

Substitute and multiply out:

\(\frac{1}{6} - \frac{5}{4}x + \frac{17}{72}x^2\).