(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:

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