(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:

\(y \approx -0.369\).