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

\(8\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) + 3 = 0\)

Simplifying, we get:

\(-1 + \frac{a}{4} - \frac{b}{2} + 3 = 0\)

\(a - 2b + 8 = 0\)

When \(p(x)\) is divided by \((2x - 1)\), substituting \(x = \frac{1}{2}\) gives a remainder of 1:

\(8\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 3 = 1\)

Simplifying, we get:

\(1 + \frac{a}{4} + \frac{b}{2} + 3 = 1\)

\(a + 2b + 12 = 0\)

Solving the equations \(a - 2b + 8 = 0\) and \(a + 2b + 12 = 0\), we find:

\(a = -10\), \(b = -1\)

(ii) With \(a = -10\) and \(b = -1\), divide \(p(x) = 8x^3 - 10x^2 - x + 3\) by \(2x^2 - 1\):

The division gives a quotient of \(4x - 5\) and a remainder of \(3x - 2\).

(i) Since \((3x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{3}\) into \(p(x)\) should yield zero:

\(a\left(-\frac{1}{3}\right)^3 - 20\left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right) + 3 = 0\)

\(-\frac{a}{27} - \frac{20}{9} - \frac{1}{3} + 3 = 0\)

Solving for \(a\):

\(-\frac{a}{27} = \frac{20}{9} + \frac{1}{3} - 3\)

\(-\frac{a}{27} = \frac{20}{9} + \frac{3}{9} - \frac{27}{9}\)

\(-\frac{a}{27} = -\frac{4}{9}\)

\(a = 12\)

(ii) With \(a = 12\), the polynomial becomes \(12x^3 - 20x^2 + x + 3\). Divide by \(3x + 1\) to find the quadratic factor:

Perform polynomial division to get \(4x^2 - 8x + 3\).

Factor the quadratic:

\(4x^2 - 8x + 3 = (2x - 1)(2x - 3)\)

Thus, the complete factorisation is \((3x+1)(2x-1)(2x-3)\).

To divide \(2x^2\) by \(x + 2\), we perform polynomial long division.

1. Divide the first term of the dividend \(2x^2\) by the first term of the divisor \(x\):

\(\frac{2x^2}{x} = 2x\)

2. Multiply the entire divisor \(x + 2\) by \(2x\):

\(2x(x + 2) = 2x^2 + 4x\)

3. Subtract \(2x^2 + 4x\) from \(2x^2\):

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

4. Divide \(-4x\) by \(x\):

\(\frac{-4x}{x} = -4\)

5. Multiply the entire divisor \(x + 2\) by \(-4\):

\(-4(x + 2) = -4x - 8\)

6. Subtract \(-4x - 8\) from \(-4x\):

\(-4x - (-4x - 8) = 8\)

The quotient is \(2x - 4\) and the remainder is 8.

(i) Since \((x - 2)\) is a factor of \(p(x)\), by the factor theorem, \(p(2) = 0\).

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

\(p(2) = 2^3 - 3a(2) + 4a = 0\)

\(8 - 6a + 4a = 0\)

\(8 - 2a = 0\)

\(2a = 8\)

\(a = 4\)

(ii)(a) Substitute \(a = 4\) into \(p(x)\):

\(p(x) = x^3 - 12x + 16\)

Divide \(p(x)\) by \((x-2)\) to find the other factor:

\(x^3 - 12x + 16 = (x-2)(x^2 + 2x - 8)\)

Factor \(x^2 + 2x - 8\):

\(x^2 + 2x - 8 = (x-2)(x+4)\)

Thus, \(p(x) = (x-2)^2(x+4)\)

(ii)(b) Substitute \(x^2\) into \(p(x)\):

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

Set \(p(x^2) = 0\):

\((x^2 - 2)^2 = 0\) gives \(x^2 = 2\) so \(x = \pm \sqrt{2}\)

\((x^2 + 4) = 0\) gives \(x^2 = -4\) so \(x = \pm 2i\)

The roots are \(\pm \sqrt{2}, \pm 2i\).

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

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

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

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

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

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

\(\Rightarrow 7a = 14\)

\(\Rightarrow a = 2\)

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

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

We know \((2x - 1)\) is a factor, so divide \(p(x)\) by \((2x - 1)\) to find the other factor:

Using polynomial division or inspection, we find:

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