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

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

Using a correct method for finding coefficients, we find:

\(A = 1\), \(B = -1\), \(C = 6\).

(b) Obtain the expansion of \(f(x)\):

Use a correct method to find the first two terms of the expansion of each partial fraction:

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

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

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

Substitute back into the partial fractions and simplify:

\(f(x) = (1 + x + x^2) + \left(-\frac{1}{2} - \frac{3}{4}x - \frac{9}{8}x^2\right) + \left(\frac{6}{4} + \frac{18}{4}x + \frac{81}{8}x^2\right)\)

Combine and simplify to get:

\(f(x) = 2 - \frac{11}{4}x + 10x^2\).

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

Assume \(\frac{2x(5-x)}{(3+x)(1-x)^2} = \frac{A}{3+x} + \frac{B}{1-x} + \frac{C}{(1-x)^2}\).

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

\(2x(5-x) = A(1-x)^2 + B(3+x)(1-x) + C(3+x)\).

Expand and equate coefficients to find \(A = -3\), \(B = -1\), \(C = 2\).

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

Use the expansions:

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

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

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

Substitute into the partial fractions:

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

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

\(\frac{2}{(1-x)^2} = 2 + 4x + 6x^2 + \cdots\)

Add the expansions and simplify:

\(f(x) = \left(-1 + \frac{1}{3}x - \frac{1}{9}x^2 + \cdots \right) + \left(-1 - x - x^2 + \cdots \right) + \left(2 + 4x + 6x^2 + \cdots \right)\)

Combine like terms:

\(f(x) = \frac{10}{3}x + \frac{44}{9}x^2 + \frac{190}{27}x^3\).