The functions f and g are defined as follows: $f : x \mapsto |e^{x-3}-1|,\quad x \le 3$ $g : x \mapsto a^2 - x^2,\quad x \in \mathbb{R},\; x < 0$, where $0 < a < 1$.

1. Simplify the expression for $f(x)$ by determining the sign of the term inside the modulus for the given domain.

Given the domain $x \le 3$, we have:
$x - 3 \le 0$
$e^{x-3} \le e^0 = 1$
$e^{x-3} - 1 \le 0$

Hence, we can remove the modulus sign by negating the expression:
$f(x) = |e^{x-3} - 1| = -(e^{x-3} - 1) = 1 - e^{x-3}$

2. Let $y = f(x)$ and make $x$ the subject.

$y = 1 - e^{x-3}$
$e^{x-3} = 1 - y$
$x - 3 = \ln(1 - y)$
$x = 3 + \ln(1 - y)$

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

Since $e^{x-3} > 0$ for all real $x$:
$1 - e^{x-3} < 1$

Also, from Step 1, $1 - e^{x-3} \ge 0$.
Therefore, the range of $f$ is $0 \le y < 1$.
This corresponds to the domain of $f^{-1}$, so its domain is $0 \le x < 1$.

4. State the inverse function using the required mapping notation.

$f^{-1} : x \mapsto 3 + \ln(1 - x), \quad 0 \le x < 1$

1. Determine the key features of the graph of $y = g(x)$.

The function is $g(x) = a^2 - x^2$ for $x < 0$. The graph is the left half of a downward-opening parabola.
As $x \to 0^-$, $y \to a^2$, so there is an open circle at $(0, a^2)$.
To find the $x$-intercept, set $y = 0$:
$a^2 - x^2 = 0 \implies x = -a \quad \text{(since } x < 0\text{)}$

2. Determine the key features of the graph of $y = g^{-1}(x)$.

The graph of $y = g^{-1}(x)$ is the reflection of $y = g(x)$ in the line $y = x$.
Reflecting the open circle $(0, a^2)$ gives an open circle at $(a^2, 0)$.
Reflecting the $x$-intercept $(-a, 0)$ gives a $y$-intercept at $(0, -a)$.
The graph opens leftwards in the third quadrant.

3. Determine the key features of the graph of $y = g^{-1}g(x)$ and synthesise the final sketch.

By definition of inverse functions, $g^{-1}g(x) = x$ for all $x$ in the domain of $g$.
Since the domain of $g$ is $x < 0$, the graph of $y = g^{-1}g(x)$ is the straight line $y = x$ restricted to $x < 0$, ending with an open circle at $(0,0)$.
Furthermore, the curves $y = g(x)$ and $y = g^{-1}(x)$ intersect each other on this line $y = x$ in the third quadrant.
Since $0 < a < 1$, we must have $a^2 < a$, so the sketch should visually reflect that $a^2$ is closer to the origin than $a$.

1. Set up the equation for the intersection by using the property of inverse functions.

Since $g'(x) = -2x > 0$ for $x < 0$, $g(x)$ is an increasing function on its domain. Therefore, the intersections of $y = g(x)$ and $y = g^{-1}(x)$ lie on the line $y = x$.
$g(x) = x$
$a^2 - x^2 = x$
$x^2 + x - a^2 = 0$

2. Solve the quadratic equation and reject the invalid root based on the domain restriction.

Using the quadratic formula,
$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-a^2)}}{2}$
$x = \frac{-1 \pm \sqrt{1 + 4a^2}}{2}$
Since the domain of $g$ is $x < 0$ and $\sqrt{1 + 4a^2} > 1$ (as $a > 0$), the positive root $x = \frac{-1 + \sqrt{1 + 4a^2}}{2}$ is outside the domain and must be rejected.
Therefore,
$x = \frac{-1 - \sqrt{1 + 4a^2}}{2}$

1. Find the range of the inner function $g$.

Given $g(x) = a^2 - x^2$ for $x < 0$.
Since $x < 0$, we have $x^2 > 0$.
Therefore, $a^2 - x^2 < a^2$, which implies $g(x) < a^2$.
As $x \to -\infty$, $g(x) \to -\infty$.
Thus, the range of $g$ is $\mathrm{R}_g = (-\infty, a^2)$.

2. Show that the composite function $fg$ exists.

The domain of $f$ is given as $\mathrm{D}_f = (-\infty, 3]$.
We are given $0 < a < 1$, which implies $a^2 < 1$.
Hence, $a^2 < 1 < 3$, which means the interval $(-\infty, a^2)$ is completely contained within $(-\infty, 3]$.
Thus, $\mathrm{R}_g \subseteq \mathrm{D}_f$.
Because the range of the inner function $g$ is a subset of the domain of the outer function $f$, the composite function $fg$ exists.

3. Find the range of $fg$ by evaluating $f$ over $\mathrm{R}_g$.

The range of $fg$ is the set of values $f(u)$ takes for $u \in \mathrm{R}_g = (-\infty, a^2)$.
For $u < a^2 < 3$, $u - 3 < 0$, which means $e^{u-3} < e^0 = 1$.
Thus, $e^{u-3} - 1 < 0$, and we can drop the absolute value by negating the expression:
$f(u) = |e^{u-3} - 1| = 1 - e^{u-3}$
As $u \to -\infty$, $e^{u-3} \to 0$, so $f(u) \to 1 - 0 = 1$.
As $u \to a^2$, $f(u) \to 1 - e^{a^2-3}$.
Since $1 - e^{u-3}$ is a strictly decreasing function of $u$, the range of $fg$ is $(1 - e^{a^2-3}, 1)$.