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

(i) To express \(f(x)\) in the form \(\frac{A}{2-x} + \frac{Bx+C}{4+x^2}\), we equate:

\(\frac{12 + 8x - x^2}{(2-x)(4+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{4+x^2}\)

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

\(12 + 8x - x^2 = A(4+x^2) + (Bx+C)(2-x)\)

Expand and collect like terms:

\(12 + 8x - x^2 = 4A + Ax^2 + 2Bx + Cx - Bx^2 - Cx\)

\(= (A-B)x^2 + (2B+C)x + 4A\)

Equating coefficients, we get:

\(A - B = -1\)

\(2B + C = 8\)

\(4A = 12\)

Solving these equations gives \(A = 3\), \(B = 4\), \(C = 0\).

(ii) To show \(\int_0^1 f(x) \, dx = \ln\left(\frac{25}{2}\right)\), integrate each term separately:

\(\int \frac{3}{2-x} \, dx = -3 \ln|2-x|\)

\(\int \frac{4x}{4+x^2} \, dx = 2 \ln|4+x^2|\)

\(\int \frac{0}{4+x^2} \, dx = 0\)

Evaluate from 0 to 1:

\(-3 \ln|2-1| + 2 \ln|4+1^2| - (-3 \ln|2-0| + 2 \ln|4+0^2|)\)

\(= -3 \ln(1) + 2 \ln(5) + 3 \ln(2) - 2 \ln(4)\)

\(= 2 \ln(5) + 3 \ln(2) - 2 \ln(4)\)

\(= \ln\left(\frac{5^2 \cdot 2^3}{4^2}\right)\)

\(= \ln\left(\frac{25}{2}\right)\)

First, express \(\frac{2x + 7}{(2x + 1)(x + 2)}\) as partial fractions:

\(\frac{2x + 7}{(2x + 1)(x + 2)} = \frac{A}{2x + 1} + \frac{B}{x + 2}\).

Multiply through by \((2x + 1)(x + 2)\) to clear the denominators:

\(2x + 7 = A(x + 2) + B(2x + 1)\).

Expanding gives:

\(2x + 7 = Ax + 2A + 2Bx + B\).

Combine like terms:

\(2x + 7 = (A + 2B)x + (2A + B)\).

Equating coefficients, we get:

\(A + 2B = 2\)

\(2A + B = 7\).

Solving these equations, we find:

\(A = 4\), \(B = -1\).

Thus, \(\frac{2x + 7}{(2x + 1)(x + 2)} = \frac{4}{2x + 1} - \frac{1}{x + 2}\).

Now integrate each term separately:

\(\int \left( \frac{4}{2x + 1} - \frac{1}{x + 2} \right) \, dx = 2 \ln |2x + 1| - \ln |x + 2| + C\).

Apply the limits from 0 to 7:

\(\left[ 2 \ln |2x + 1| - \ln |x + 2| \right]_0^7\).

Calculate at the upper limit (7):

\(2 \ln |15| - \ln |9| = 2 \ln 15 - \ln 9\).

Calculate at the lower limit (0):

\(2 \ln |1| - \ln |2| = 0 - \ln 2\).

Subtract the lower limit from the upper limit:

\((2 \ln 15 - \ln 9) - (0 - \ln 2) = 2 \ln 15 - \ln 9 + \ln 2\).

Simplify using logarithm properties:

\(2 \ln 15 - \ln 9 + \ln 2 = \ln \left( \frac{15^2 \cdot 2}{9} \right) = \ln 50\).

(i) Express \(\frac{2}{(x+1)(x+3)}\) as \(\frac{A}{x+1} + \frac{B}{x+3}\).

Multiply through by \((x+1)(x+3)\) to get: \(2 = A(x+3) + B(x+1)\).

Set \(x = -1\): \(2 = A(-1+3) \Rightarrow A = 1\).

Set \(x = -3\): \(2 = B(-3+1) \Rightarrow B = -1\).

Thus, \(\frac{2}{(x+1)(x+3)} = \frac{1}{x+1} - \frac{1}{x+3}\).

(ii) Square the result from part (i):

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

Substitute \(\frac{2}{(x+1)(x+3)} = \frac{1}{x+1} - \frac{1}{x+3}\) to get:

\(\frac{1}{(x+1)^2} - \frac{1}{x+1} + \frac{1}{x+3} + \frac{1}{(x+3)^2}\).

(iii) Integrate \(\int_0^1 \frac{4}{(x+1)^2(x+3)^2} \, dx\).

Using the result from part (ii):

\(\int_0^1 \left( \frac{1}{(x+1)^2} - \frac{1}{x+1} + \frac{1}{x+3} + \frac{1}{(x+3)^2} \right) \, dx\).

Integrate each term:

\(\int \frac{1}{(x+1)^2} \, dx = -\frac{1}{x+1}\), \(\int \frac{1}{x+1} \, dx = \ln(x+1)\), \(\int \frac{1}{x+3} \, dx = \ln(x+3)\), \(\int \frac{1}{(x+3)^2} \, dx = -\frac{1}{x+3}\).

Evaluate from 0 to 1:

\(-\frac{1}{2} - \ln 2 + \ln 4 - \frac{1}{4} = \frac{7}{12} - \ln \frac{3}{2}\).

To solve \(\int_1^2 \frac{2}{u(4-u)} \, du\), we use partial fraction decomposition.

Express \(\frac{2}{u(4-u)}\) as \(\frac{A}{u} + \frac{B}{4-u}\).

Equating coefficients, we have:

\(2 = A(4-u) + Bu\)

\(2 = 4A - Au + Bu\)

\(2 = 4A + (B-A)u\)

Comparing coefficients, we get:

\(4A = 2\) and \(B-A = 0\)

Solving these, \(A = \frac{1}{2}\) and \(B = \frac{1}{2}\).

Thus, \(\frac{2}{u(4-u)} = \frac{1/2}{u} + \frac{1/2}{4-u}\).

Integrate each term separately:

\(\int \frac{1/2}{u} \, du = \frac{1}{2} \ln |u|\)

\(\int \frac{1/2}{4-u} \, du = -\frac{1}{2} \ln |4-u|\)

Evaluate from 1 to 2:

\(\left[ \frac{1}{2} \ln |u| - \frac{1}{2} \ln |4-u| \right]_1^2\)

\(= \left( \frac{1}{2} \ln 2 - \frac{1}{2} \ln 2 \right) - \left( \frac{1}{2} \ln 1 - \frac{1}{2} \ln 3 \right)\)

\(= 0 - (-\frac{1}{2} \ln 3)\)

\(= \frac{1}{2} \ln 3\)