(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.

(i) To find stationary points, we differentiate \(y = x^3 + ax^2 + bx\) to get \(\frac{dy}{dx} = 3x^2 + 2ax + b\). For no stationary points, the discriminant of this quadratic must be less than zero: \(b^2 - 4ac < 0\). Here, \(a = 2a\) and \(c = b\), so \(b^2 - 4(3)(b) < 0\) simplifies to \(a^2 < 3b\).

(ii) With \(a = -6\) and \(b = 9\), the equation becomes \(y = x^3 - 6x^2 + 9x\). Differentiating gives \(\frac{dy}{dx} = 3x^2 - 12x + 9\). Setting \(\frac{dy}{dx} < 0\) gives \(3x^2 - 12x + 9 < 0\). Solving \(3(x^2 - 4x + 3) < 0\) gives roots at \(x = 1\) and \(x = 3\). Thus, \(y\) is decreasing for \(1 < x < 3\).

(i) To find \(f'(x)\), use the quotient rule: if \(f(x) = \frac{u}{v}\), then \(f'(x) = \frac{u'v - uv'}{v^2}\).

Here, \(u = 5\) and \(v = 1 - 3x\).

\(u' = 0\) and \(v' = -3\).

So, \(f'(x) = \frac{0 imes (1 - 3x) - 5 imes (-3)}{(1 - 3x)^2} = \frac{15}{(1 - 3x)^2}\).

Since the derivative is negative, \(f'(x) = \frac{-15}{(1 - 3x)^2}\).

(ii) Since \(15 > 0\) and \((1 - 3x)^2 > 0\) for \(x \geq 1\), \(f'(x) > 0\).

Therefore, \(f\) is an increasing function.

To determine if \(f(x) = (2x - 5)^3 + x\) is increasing, we need to find its derivative \(f'(x)\).

Using the chain rule, the derivative of \((2x - 5)^3\) is \(3(2x - 5)^2 \times 2\).

Thus, \(f'(x) = 3(2x - 5)^2 \times 2 + 1 = 6(2x - 5)^2 + 1\).

We can also express this as \(f'(x) = 24\left(\frac{x - 5}{2}\right)^2 + 1\).

Since \((2x - 5)^2 \geq 0\) for all \(x\), it follows that \(6(2x - 5)^2 + 1 > 0\) for all \(x\).

Therefore, \(f'(x) > 0\) for all \(x \in \mathbb{R}\), which means \(f\) is an increasing function.

To determine if \(f(x) = \frac{1}{x^3} - x^3\) is a decreasing function, we need to find its derivative \(f'(x)\) and check if it is negative for \(x > 0\).

First, differentiate \(f(x)\):

\(f'(x) = \frac{d}{dx} \left( \frac{1}{x^3} \right) - \frac{d}{dx} (x^3)\).

Using the power rule, \(\frac{d}{dx} \left( x^n \right) = nx^{n-1}\), we find:

\(\frac{d}{dx} \left( \frac{1}{x^3} \right) = \frac{d}{dx} (x^{-3}) = -3x^{-4}\).

\(\frac{d}{dx} (x^3) = 3x^2\).

Thus, \(f'(x) = -3x^{-4} - 3x^2\).

Simplifying, \(f'(x) = -\frac{3}{x^4} - 3x^2\).

For \(x > 0\), both terms \(-\frac{3}{x^4}\) and \(-3x^2\) are negative, so \(f'(x) < 0\).

Therefore, \(f(x)\) is a decreasing function for \(x > 0\).