(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:

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