(i) Express \(f(x)\) as partial fractions:

Assume \(\frac{3x^2 + x + 6}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+4}\).

Multiply through by \((x+2)(x^2+4)\) to clear the denominators:

\(3x^2 + x + 6 = A(x^2+4) + (Bx+C)(x+2)\).

Expand and collect like terms:

\(3x^2 + x + 6 = Ax^2 + 4A + Bx^2 + 2Bx + Cx + 2C\).

Combine terms:

\(3x^2 + x + 6 = (A+B)x^2 + (2B+C)x + (4A+2C)\).

Equate coefficients:

\(A + B = 3\)

\(2B + C = 1\)

\(4A + 2C = 6\)

Solve these equations to find \(A = 2\), \(B = 1\), \(C = -1\).

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

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the binomial expansion for \((x+2)^{-1}\) and \((x^2+4)^{-1}\):

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

\((x^2+4)^{-1} = \frac{1}{4} - \frac{1}{16}x^2 + \cdots\)

Substitute back into the partial fractions:

\(\frac{2}{x+2} = 2 \left( \frac{1}{2} - \frac{1}{4}x \right) = 1 - \frac{1}{2}x\)

\(\frac{x-1}{x^2+4} = (x-1) \left( \frac{1}{4} - \frac{1}{16}x^2 \right) = \frac{1}{4}x - \frac{1}{4} - \frac{1}{16}x^3 + \cdots\)

Add the expansions:

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

Combine like terms:

\(f(x) = \frac{3}{4} - \frac{1}{4}x + \frac{5}{16}x^2\).