The functions $\text{f}$ and $\text{g}$ are defined by
$\text{f} : x \mapsto \frac{ax - 6}{x - 3} \quad \text{for } x \in \mathbb{R}, \ x \ne 3,$
$\text{g} : x \mapsto \text{e}^{-x} \quad \text{for } x \in \mathbb{R}, \ x \ge \ln 3.$
The function $\text{f}$ is such that $\text{f}(x) = \text{f}^{-1}(x)$ for all $x$ in the domain of $\text{f}$.

1. Set up an equation to find the inverse function $\text{f}^{-1}(x)$.

Let $y = \frac{ax - 6}{x - 3}$.

2. Rearrange the equation to make $x$ the subject.

$y(x - 3) = ax - 6$
$xy - 3y = ax - 6$
$xy - ax = 3y - 6$
$x(y - a) = 3y - 6$
$x = \frac{3y - 6}{y - a}$

Therefore, $\text{f}^{-1}(x) = \frac{3x - 6}{x - a}$.

3. Use the given condition $\text{f}(x) = \text{f}^{-1}(x)$ to find $a$.

Given $\text{f}(x) = \text{f}^{-1}(x)$ for all $x$ in the domain:
$\frac{ax - 6}{x - 3} = \frac{3x - 6}{x - a}$

Comparing the numerators and denominators of both sides, we can deduce that:
$a = 3$

1. Deduce the value of $\text{f}^6(\pi)$ using the given self-inverse property.

Given $\text{f}(x) = \text{f}^{-1}(x)$, applying $\text{f}$ to both sides yields:
$\text{f}^2(x) = \text{f}(\text{f}^{-1}(x)) = x$
It follows that any even number of compositions of $\text{f}$ will return $x$:
$\text{f}^6(x) = x$
Substituting $x = \pi$:
$\text{f}^6(\pi) = \pi$

1. Determine the function $\text{f}(x)$ using the context from part (a)

From part (a), let $y = \frac{ax - 6}{x - 3}$. Rearranging for $x$:
$xy - 3y = ax - 6$
$x(y - a) = 3y - 6 \implies x = \frac{3y - 6}{y - a}$
So $\text{f}^{-1}(x) = \frac{3x - 6}{x - a}$.
Since $\text{f}(x) = \text{f}^{-1}(x)$ for all $x$ in the domain, we match coefficients to get $a = 3$.
Thus, $\text{f}(x) = \frac{3x - 6}{x - 3}$.

2. Find the exact range of the inner function $\text{g}(x)$

The domain of $\text{g}$ is $x \ge \ln 3$.
Since $\text{g}(x) = \text{e}^{-x}$ is a strictly decreasing function and $\text{e}^{-x} > 0$ for all real $x$:
$0 < \text{e}^{-x} \le \text{e}^{-\ln 3}$
$0 < \text{g}(x) \le \frac{1}{3}$
So the exact range of $\text{g}$ is $\left(0, \frac{1}{3}\right]$.

3. Evaluate $\text{f}$ over the range of $\text{g}$ to find the exact range of $\text{fg}$

The range of $\text{g}$ becomes the domain for $\text{f}$. We evaluate $\text{f}(x)$ at the boundaries of $\left(0, \frac{1}{3}\right]$:
As $x \to 0^+$, $\text{f}(x) \to \frac{3(0) - 6}{0 - 3} = 2$.
At $x = \frac{1}{3}$, $\text{f}\left(\frac{1}{3}\right) = \frac{3(\frac{1}{3}) - 6}{\frac{1}{3} - 3} = \frac{1 - 6}{-\frac{8}{3}} = \frac{15}{8}$.
To confirm $\text{f}(x)$ is monotonic on this interval, we can rewrite it:
$\text{f}(x) = \frac{3(x - 3) + 3}{x - 3} = 3 + \frac{3}{x - 3}$
For $x \in \left(0, \frac{1}{3}\right]$, $x - 3$ is negative and increasing (towards $0$), so $\frac{3}{x - 3}$ is decreasing. Thus, $\text{f}(x)$ is strictly decreasing on this interval.
The supremum approaches $2$ (exclusive) and the minimum value is $\frac{15}{8}$ (inclusive).
Therefore, the exact range of $\text{fg}$ is $\left[ \frac{15}{8}, 2 \right)$.