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

Equating coefficients, solve for \(A\), \(B\), and \(C\):

\(A = 1\), \(B = 2\), \(C = -4\).

Thus, \(\frac{2 - x + 8x^2}{(1-x)(1+2x)(2+x)} = \frac{1}{1-x} + \frac{2}{1+2x} - \frac{4}{2+x}\).

(ii) Expand each partial fraction:

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

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

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

Combine the expansions up to \(x^2\):

\(1 + x + x^2 + 2 - 4x + 8x^2 - 2 + x - \frac{x^2}{2} = 1 - 2x + \frac{17}{2}x^2\).