(i) To find the value of \(a\), we divide \(p(x) = x^4 + 3x^3 + ax + 3\) by \(x^2 - x + 1\). Performing polynomial division, we obtain a quotient of \(x^2 + 4x + 3\) and a remainder of zero, which confirms divisibility. Solving for \(a\), we find \(a = 1\).

(ii) With \(a = 1\), the polynomial becomes \(x^4 + 3x^3 + x + 3\). Factoring or solving this equation, we find the real roots are \(x = -3\) and \(x = -1\).

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

\(\frac{1}{8}a + \frac{1}{4}b + \frac{5}{2} - 2 = 0\)

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

\(\Rightarrow a + 2b + 4 = 0\) (Equation 1)

When \(p(x)\) is divided by \((x - 2)\), the remainder is 12. Substituting \(x = 2\) into \(p(x)\) gives:

\(8a + 4b + 10 - 2 = 12\)

\(\Rightarrow 8a + 4b + 8 = 12\)

\(\Rightarrow 8a + 4b = 4\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(a = -2b - 4\)

Substitute into Equation 2:

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

\(-16b - 32 + 4b = 4\)

\(-12b = 36\)

\(b = -3\)

Substitute \(b = -3\) into Equation 1:

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

\(a - 6 + 4 = 0\)

\(a = 2\)

(ii) With \(a = 2\) and \(b = -3\), the polynomial is \(2x^3 - 3x^2 + 5x - 2\).

Divide \(2x^3 - 3x^2 + 5x - 2\) by \(2x - 1\) to find the quadratic factor:

The division gives a quotient of \(x^2 - x + 2\).

Thus, the quadratic factor is \(x^2 - x + 2\).

To divide \(2x^4 - 27\) by \(x^2 + x + 3\), we perform polynomial long division:

1. Divide the leading term \(2x^4\) by \(x^2\) to get \(2x^2\).

2. Multiply \(2x^2\) by \(x^2 + x + 3\) to get \(2x^4 + 2x^3 + 6x^2\).

3. Subtract \(2x^4 + 2x^3 + 6x^2\) from \(2x^4 - 27\) to get \(-2x^3 - 6x^2 - 27\).

4. Divide \(-2x^3\) by \(x^2\) to get \(-2x\).

5. Multiply \(-2x\) by \(x^2 + x + 3\) to get \(-2x^3 - 2x^2 - 6x\).

6. Subtract \(-2x^3 - 2x^2 - 6x\) from \(-2x^3 - 6x^2 - 27\) to get \(-4x^2 - 6x - 27\).

7. Divide \(-4x^2\) by \(x^2\) to get \(-4\).

8. Multiply \(-4\) by \(x^2 + x + 3\) to get \(-4x^2 - 4x - 12\).

9. Subtract \(-4x^2 - 4x - 12\) from \(-4x^2 - 6x - 27\) to get \(-2x - 15\).

The quotient is \(2x^2 - 2x - 4\) and the remainder is \(10x - 15\).

(i) To show that \(f(-2) = 0\), substitute \(x = -2\) into \(f(x)\):

\(f(-2) = 12(-2)^3 + 25(-2)^2 - 4(-2) - 12\)

\(= -96 + 100 + 8 - 12\)

\(= 0\)

Thus, \(x + 2\) is a factor of \(f(x)\).

To factorise \(f(x)\) completely, divide \(f(x)\) by \(x + 2\):

Using synthetic division or polynomial division, we find:

\(12x^3 + 25x^2 - 4x - 12 = (x+2)(12x^2 + x - 6)\)

Factor the quadratic \(12x^2 + x - 6\):

\(12x^2 + x - 6 = (4x+3)(3x-2)\)

Thus, \(f(x) = (x+2)(4x+3)(3x-2)\).

(ii) Given \(12 \times 27^y + 25 \times 9^y - 4 \times 3^y - 12 = 0\), let \(3^y = k\).

Then \(9^y = k^2\) and \(27^y = k^3\).

Substitute into the equation:

\(12k^3 + 25k^2 - 4k - 12 = 0\)

By inspection or trial, \(k = \frac{2}{3}\) satisfies the equation.

Thus, \(3^y = \frac{2}{3}\).

Taking logarithms, \(y = \log_3 \left( \frac{2}{3} \right)\).

Calculate \(y\) to 3 significant figures:

\(y \approx -0.369\).

(i) Since \((z + 2)\) is a factor of \(p(z)\), we have \(p(-2) = 0\).

Substitute \(z = -2\) into \(p(z)\):

\((-2)^3 + m(-2)^2 + 24(-2) + 32 = 0\)

\(-8 + 4m - 48 + 32 = 0\)

\(4m - 24 = 0\)

\(4m = 24\)

\(m = 6\)

(ii)(a) With \(m = 6\), the polynomial becomes \(p(z) = z^3 + 6z^2 + 24z + 32\).

Since \(z = -2\) is a root, divide \(p(z)\) by \((z + 2)\) to find the quadratic factor:

\(z^3 + 6z^2 + 24z + 32 = (z + 2)(z^2 + 4z + 16)\)

Solving \(z^2 + 4z + 16 = 0\) using the quadratic formula:

\(z = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}\)

\(z = \frac{-4 \pm \sqrt{-48}}{2}\)

\(z = \frac{-4 \pm 4\sqrt{3}i}{2}\)

\(z = -2 \pm 2\sqrt{3}i\)

Thus, the roots are \(z = -2, -2 + 2\sqrt{3}i, -2 - 2\sqrt{3}i\).

(ii)(b) To find the roots of \(p(z^2) = 0\), substitute \(z^2\) into the roots found in part (ii)(a):

For \(z = -2\), \(z^2 = 4\), so \(z = \pm i\sqrt{2}\).

For \(z = -2 + 2\sqrt{3}i\), find the square roots:

Let \(z = a + bi\), then \(a^2 - b^2 = -2\) and \(2ab = 2\sqrt{3}\).

Solving gives \(a = 1, b = \sqrt{3}\) or \(a = -1, b = -\sqrt{3}\).

Thus, \(z = 1 + i\sqrt{3}, 1 - i\sqrt{3}, -1 + i\sqrt{3}, -1 - i\sqrt{3}\).

The six roots of \(p(z^2) = 0\) are \(\pm i\sqrt{2}, \pm (1+i\sqrt{3}), \pm (1-i\sqrt{3})\).