(i) We express \(\frac{4 + 12x + x^2}{(3-x)(1+2x)^2}\) in partial fractions. Assume the form:

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

Equating coefficients, we find:

\(A = 1, \ B = \frac{3}{2}, \ C = -\frac{1}{2}\)

(ii) To expand \(\frac{4 + 12x + x^2}{(3-x)(1+2x)^2}\), we first expand \((3-x)^{-1}\) and \((1+2x)^{-2}\):

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

\((1+2x)^{-2} = 1 - 4x + 8x^2 + \cdots\)

Using these expansions, we find the expansion of each partial fraction up to \(x^2\):

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

Combining these, we obtain:

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

(i) To express \(\frac{7x^2 + 8}{(1+x)^2(2-3x)}\) in partial fractions, assume the form:

\(\frac{D}{(1+x)^2} + \frac{E}{1+x} + \frac{F}{2-3x}\)

Equating and solving for constants, we find:

\(D = -1\), \(E = 2\), \(F = 4\).

Thus, the partial fraction decomposition is:

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

(ii) To expand \(\frac{7x^2 + 8}{(1+x)^2(2-3x)}\), use the expansions:

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

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

Combine these expansions to find the first three terms:

\(4 - 2x + \frac{25}{2}x^2\).

(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:

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

(i) Express \(\frac{9 - 7x + 8x^2}{(3-x)(1+x^2)}\) in the form \(\frac{A}{3-x} + \frac{Bx+C}{1+x^2}\).

Equating coefficients, we find:

\(A = 6\), \(B = -2\), \(C = 1\).

(ii) Expand \(\frac{1}{3-x}\) as \(\frac{1}{3} \left( 1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} \right)\).

Expand \(\frac{Bx+C}{1+x^2}\) as \((Bx+C)(1-x^2)\).

Combine the expansions to obtain:

\(3 - \frac{4}{3}x - \frac{7}{9}x^2 + \frac{56}{27}x^3\).

(i) Express \(\frac{5x - x^2}{(1+x)(2+x^2)}\) in partial fractions:

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

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

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

Expand and collect like terms:

\(5x - x^2 = 2A + Ax^2 + Bx + Bx^2 + C + Cx\).

Equate coefficients:

For \(x^2\): \(-1 = A + B\)

For \(x\): \(5 = B + C\)

For constant term: \(0 = 2A + C\)

Solve the system of equations:

1. \(A + B = -1\)

2. \(B + C = 5\)

3. \(2A + C = 0\)

From equation 3: \(C = -2A\).

Substitute \(C = -2A\) into equation 2: \(B - 2A = 5\).

Substitute \(B = -1 - A\) from equation 1 into \(B - 2A = 5\):

\(-1 - A - 2A = 5\)

\(-1 - 3A = 5\)

\(-3A = 6\)

\(A = -2\)

Substitute \(A = -2\) into \(B = -1 - A\):

\(B = -1 + 2 = 1\)

Substitute \(A = -2\) into \(C = -2A\):

\(C = 4\)

Thus, \(\frac{5x - x^2}{(1+x)(2+x^2)} = \frac{-2}{1+x} + \frac{x + 4}{2+x^2}\).

(ii) Expand \(\frac{5x - x^2}{(1+x)(2+x^2)}\) in ascending powers of \(x\):

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

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

Expand \(\frac{x + 4}{2+x^2}\) using the binomial series:

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

Combine the expansions:

\(-2 + 2x - 2x^2 + 2x^3 + \frac{x}{2} + 2 - \frac{x^3}{4} + \cdots\)

Combine like terms:

\(\frac{5}{2}x - 3x^2 + \frac{7}{4}x^3 + \cdots\)