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

Assume \(\frac{8x^2 + 9x + 8}{(1-x)(2x+3)^2} = \frac{A}{1-x} + \frac{B}{2x+3} + \frac{C}{(2x+3)^2}\).

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

\(8x^2 + 9x + 8 = A(2x+3)^2 + B(1-x)(2x+3) + C(1-x)\).

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

\(A = 1, B = -2, C = 5\).

Thus, \(f(x) = \frac{1}{1-x} - \frac{2}{2x+3} + \frac{5}{(2x+3)^2}\).

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

Use the binomial expansion for each term:

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

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

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

Combine the expansions:

\(f(x) = \left(1 + x + x^2 + \cdots\right) - 2\left(\frac{1}{3} - \frac{2}{9}x + \cdots\right) + 5\left(\frac{1}{9} - \frac{4}{27}x + \cdots\right)\).

Simplify to obtain the expansion up to \(x^2\):

\(f(x) = \frac{8}{9} + \frac{19}{27}x + \frac{13}{9}x^2\).