(i) To express \(9x^2 - 12x + 5\) in the form \((ax + b)^2 + c\), we complete the square:

Start with \(9x^2 - 12x\).

Factor out the 9: \(9(x^2 - \frac{4}{3}x)\).

Complete the square inside the bracket: \(x^2 - \frac{4}{3}x = (x - \frac{2}{3})^2 - (\frac{2}{3})^2\).

Thus, \(9(x^2 - \frac{4}{3}x) = 9((x - \frac{2}{3})^2 - \frac{4}{9}) = 9(x - \frac{2}{3})^2 - 4\).

So, \(9x^2 - 12x + 5 = 9(x - \frac{2}{3})^2 - 4 + 5 = 9(x - \frac{2}{3})^2 + 1\).

Therefore, \((3x - 2)^2 + 1\).

(ii) To determine if \(3x^3 - 6x^2 + 5x - 12\) is increasing, decreasing, or neither, find the derivative:

\(f'(x) = 9x^2 - 12x + 5\).

From part (i), \(f'(x) = (3x - 2)^2 + 1\).

Since \((3x - 2)^2\) is always non-negative, \((3x - 2)^2 + 1 > 0\) for all \(x\).

Therefore, \(f'(x) > 0\) for all \(x\), indicating that the function is always increasing.