函数 f 和 g 定义为 $f : x \mapsto [\ln(x-1)]^{2} + 2,\quad x \ge a,$ $g : x \mapsto 4 + 3x - x^{2},\quad x \le \frac{3}{2}.$

1. Find the turning point of $f(x)$ to determine the boundary for a one-to-one domain.

For $f^{-1}$ to exist, $f$ must be a one-to-one function.
Observe that $f(x) = [\ln(x-1)]^2 + 2$ has a squared term. Its turning point (minimum) occurs when the term being squared is zero:
$\ln(x-1) = 0$
$x - 1 = e^0 = 1 \implies x = 2$

2. Deduce the smallest value of $a$ for the restricted domain.

For $f(x)$ to be one-to-one on the domain $x \ge a$, the interval must not cross the turning point. This means the domain must start at or after the turning point.
Therefore, $a \ge 2$.
The smallest possible value of $a$ is $2$.

1. Let $y = g(x)$ and rearrange into a quadratic equation to solve for $x$ in terms of $y$.

Let $y = 4 + 3x - x^2$
$x^2 - 3x + (y - 4) = 0$

Using the quadratic formula to solve for $x$:
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(y - 4)}}{2(1)}$
$x = \frac{3 \pm \sqrt{9 - 4y + 16}}{2}$
$x = \frac{3 \pm \sqrt{25 - 4y}}{2}$

2. Use the given domain of $g$ to choose the correct root and state the expression for $g^{-1}(x)$.

Since the domain of $g$ is $x \le \frac{3}{2}$, we reject the positive root (which would give $x \ge \frac{3}{2}$).

$x = \frac{3 - \sqrt{25 - 4y}}{2}$

Therefore, replacing $y$ with $x$, the inverse function is:
$g^{-1}(x) = \frac{3 - \sqrt{25 - 4x}}{2}$

3. Find the range of $g$ to determine the domain of $g^{-1}$.

The domain of $g^{-1}$ is the range of $g$.
The turning point of the parabola $y = 4 + 3x - x^2$ occurs at $x = -\frac{3}{2(-1)} = \frac{3}{2}$.
Maximum value of $g = 4 + 3\left(\frac{3}{2}\right) - \left(\frac{3}{2}\right)^2 = 4 + \frac{9}{2} - \frac{9}{4} = \frac{25}{4}$.

Given the domain of $g$ is $x \le \frac{3}{2}$, the curve takes all values up to and including the maximum. Thus, the range of $g$ is $g(x) \le \frac{25}{4}$.

Hence, the domain of $g^{-1}$ is $x \le \frac{25}{4}$.

1. Find the value of $a$ from part (i) to determine the range of $f^{-1}$.

$f'(x) = \frac{2\ln(x-1)}{x-1}$
For $f$ to have an inverse, it must be a one-to-one function. The turning point is at $x = 2$, so the smallest value of $a$ is $2$.
The range of an inverse function is the domain of its original function, so $\text{R}_{f^{-1}} = \text{D}_f = [2, \infty)$.

2. Find the domain of $g^{-1}$ from part (ii).

Completing the square for $g(x)$:
$g(x) = 4 + 3x - x^2 = \frac{25}{4} - \left(x - \frac{3}{2}\right)^2$
Given the domain of $g$ is $x \le \frac{3}{2}$, the maximum value is $\frac{25}{4}$, giving the range of $g$ as $\left(-\infty, \frac{25}{4}\right]$.
The domain of $g^{-1}$ is the range of $g$, so $\text{D}_{g^{-1}} = \text{R}_g = \left(-\infty, \frac{25}{4}\right]$.

3. Evaluate the condition for the existence of $g^{-1} \circ f^{-1}$ and conclude.

For the composite function $g^{-1} \circ f^{-1}$ to exist, we require the range of the inner function to be a subset of the domain of the outer function:
$\text{R}_{f^{-1}} \subseteq \text{D}_{g^{-1}}$
However, $[2, \infty) \not\subseteq \left(-\infty, \frac{25}{4}\right]$ (for example, $7 \in \text{R}_{f^{-1}}$ but $7 \notin \text{D}_{g^{-1}}$).
Therefore, the composite function $g^{-1} \circ f^{-1}$ does not exist.