To determine whether the function \(f(x) = \frac{1}{3x+2} + x^2\) is increasing or decreasing, we need to find its derivative \(f'(x)\).

The derivative of \(f(x)\) is given by:

\(f'(x) = \left[ -\left(3x+2\right)^{-2} \right] \times [3] + [2x]\)

Simplifying, we have:

\(f'(x) = -\frac{3}{(3x+2)^2} + 2x\)

For \(x < -1\), the term \(-\frac{3}{(3x+2)^2}\) is negative and dominates over \(2x\), making \(f'(x) < 0\).

Since \(f'(x) < 0\) for \(x < -1\), the function \(f\) is decreasing.

To find the least possible value of \(n\), we need \(f'(x) \geq 0\) for \(x > n\) because the function is increasing.

First, solve \(f'(x) = x^2 - 6x + 8 = 0\).

Factor the quadratic: \((x - 2)(x - 4) = 0\).

The roots are \(x = 2\) and \(x = 4\).

The quadratic \(x^2 - 6x + 8\) is positive for \(x < 2\) and \(x > 4\).

Since the function is increasing for \(x > n\), the least integer \(n\) such that \(f'(x) \geq 0\) is \(n = 4\).

First, find the critical points by setting \(f'(x) = 0\):

\(x^3 + 2x^2 - 4x + 7 = 0\).

Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = 3x^2 + 4x - 4\).

Find the roots of \(f''(x)\) to determine intervals of increase/decrease:

\(3x^2 + 4x - 4 = 0\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find:

\(x = \frac{-4 \pm \sqrt{16 + 48}}{6} = \frac{-4 \pm 8}{6}\).

This gives \(x = \frac{2}{3}\) and \(x = -2\).

Test intervals around these points:

For \(-2 < x < \frac{2}{3}\), \(f'(x) < 0\).

For \(x > \frac{2}{3}\), \(f'(x) > 0\).

Since \(f'(x)\) changes sign, \(f\) is neither increasing nor decreasing.

(i) Differentiate the curve: \(\frac{dy}{dx} = 3x^2 - 18x + 24\).

Since the tangent is parallel to the line \(y = 2 - 3x\), the slope of the tangent is \(-3\).

Set \(3x^2 - 18x + 24 = -3\).

Solve for \(x\):

\(3x^2 - 18x + 24 + 3 = 0\)

\(3x^2 - 18x + 27 = 0\)

\(x^2 - 6x + 9 = 0\)

\((x - 3)^2 = 0\)

\(x = 3\)

Substitute \(x = 3\) into the original equation to find \(y\):

\(y = 3^3 - 9(3)^2 + 24(3) - 12 = 27 - 81 + 72 - 12 = 6\)

The point \(A\) is \((3, 6)\).

The equation of the tangent is \(y - 6 = -3(x - 3)\).

Simplify to get \(y = -3x + 15\).

(ii) Differentiate \(f(x)\): \(f'(x) = 3x^2 - 18x + 24\).

Set \(f'(x) > 0\) for \(f(x)\) to be increasing:

\(3(x^2 - 6x + 8) > 0\)

Factorize: \(3(x-2)(x-4) > 0\)

The critical points are \(x = 2\) and \(x = 4\).

For \(f(x)\) to be increasing, \(x > 4\).

Thus, the smallest value of \(k\) is 4.

To find when the function \(f(x) = (2x - 1)^{\frac{3}{2}} - 6x\) is decreasing, we need to find when its derivative \(f'(x)\) is less than or equal to zero.

First, find the derivative:

\(f'(x) = \left( \frac{3}{2} \right)(2x - 1)^{\frac{1}{2}} \times [2] - [6]\)

Simplifying, we get:

\(f'(x) = 3(2x - 1)^{\frac{1}{2}} - 6\)

Set \(f'(x) \leq 0\):

\(3(2x - 1)^{\frac{1}{2}} - 6 \leq 0\)

\(3(2x - 1)^{\frac{1}{2}} \leq 6\)

\((2x - 1)^{\frac{1}{2}} \leq 2\)

Square both sides:

\(2x - 1 \leq 4\)

\(2x \leq 5\)

\(x \leq \frac{5}{2}\)

Thus, the largest value of \(k\) for which \(f\) is a decreasing function is \(\frac{5}{2}\).