(i) Since \((2x - 1)\) is a factor of \(p(x)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) should yield zero:

\(p\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - a = 0\)

\(\Rightarrow \frac{a}{8} - \frac{1}{4} + 2 - a = 0\)

\(\Rightarrow \frac{a}{8} - a = -\frac{7}{4}\)

\(\Rightarrow -\frac{7a}{8} = -\frac{7}{4}\)

\(\Rightarrow a = 2\)

Substitute \(a = 2\) back into \(p(x)\):

\(p(x) = 2x^3 - x^2 + 4x - 2\)

Factorise \(p(x)\) using \((2x - 1)\):

\(p(x) = (2x - 1)(x^2 + 2)\)

(ii) Express \(\frac{8x - 13}{p(x)}\) in partial fractions:

\(\frac{8x - 13}{(2x-1)(x^2+2)} = \frac{A}{2x-1} + \frac{Bx + C}{x^2+2}\)

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

\(8x - 13 = A(x^2 + 2) + (Bx + C)(2x - 1)\)

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

\(A = -4\), \(B = 2\), \(C = 5\)

Thus, \(\frac{8x - 13}{p(x)} = \frac{-4}{2x-1} + \frac{2x + 5}{x^2 + 2}\)