(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 \(\pm i\sqrt{2}, \pm (1+i\sqrt{3}), \pm (1-i\sqrt{3})\).