(i) To express \(\frac{10}{(2-x)(1+x^2)}\) in partial fractions, assume \(\frac{10}{(2-x)(1+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{1+x^2}\).

Multiply through by \((2-x)(1+x^2)\) to get:

\(10 = A(1+x^2) + (Bx+C)(2-x)\).

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

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

Thus, \(\frac{10}{(2-x)(1+x^2)} = \frac{2}{2-x} + \frac{2x+4}{1+x^2}\).

(ii) For the expansion, consider \(\frac{2}{2-x} = \frac{1}{1-\frac{x}{2}}\) and expand using the geometric series:

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

For \(\frac{2x+4}{1+x^2}\), expand \(\frac{1}{1+x^2} = 1 - x^2 + x^4 - \ldots\) and multiply by \(2x+4\):

\((2x+4)(1 - x^2) = 2x + 4 - 2x^3 - 4x^2\).

Combine the expansions:

\(5 + \frac{5}{2}x - \frac{15}{4}x^2 - \frac{15}{8}x^3\).