The functions f and g are defined by $f : x \mapsto |4+2x - x^{2}|, \; x \in \mathbb{R}, \; x \ge 3.5,$ $g : x \mapsto 4+e^{ax}, \; x \in \mathbb{R}, \; x \ge -1,$ where $a>0$.

1. Remove the absolute value signs by determining the sign of the expression for the given domain.

By completing the square,
$4 + 2x - x^2 = 5 - (x - 1)^2$
For the given domain $x \ge 3.5$,
$5 - (x - 1)^2 \le 5 - (3.5 - 1)^2 = 5 - 6.25 = -1.25$
Since the expression inside the absolute value is always negative for $x \ge 3.5$, we negate it to remove the modulus:
$f(x) = -(4 + 2x - x^2) = x^2 - 2x - 4 = (x - 1)^2 - 5$

2. Find the rule for the inverse function by letting $y = f(x)$ and making $x$ the subject.

Let $y = f(x) = (x - 1)^2 - 5$.
$y + 5 = (x - 1)^2$
$x - 1 = \pm\sqrt{y + 5}$
Since $x \ge 3.5$, we have $x - 1 \ge 2.5 > 0$, so we must reject the negative root.
$x = 1 + \sqrt{y + 5}$
$\therefore f^{-1}(x) = 1 + \sqrt{x + 5}$

3. Determine the domain of $f^{-1}$ from the range of $f$.

The domain of $f^{-1}$ is exactly the range of $f$.
Since $f(x) = (x - 1)^2 - 5$ is an increasing function for $x \ge 3.5$, its minimum value occurs at the lower bound of the domain:
$f(3.5) = (3.5 - 1)^2 - 5 = 6.25 - 5 = 1.25$
As $x \to \infty$, $f(x) \to \infty$.
Thus, the range of $f$ is $[1.25, \infty)$.
Therefore, the domain of $f^{-1}$ is $x \ge 1.25$.

1. Use the property that intersections of a function and its inverse lie on the line $y = x$ to set up the equation.

For $x \ge 3.5$, the expression $4 + 2x - x^2 \le -1.25 < 0$, so we remove the modulus sign by negating it:
$f(x) = x^2 - 2x - 4$
Since $f$ is a strictly increasing function on this domain, any intersection of $y = f(x)$ and $y = f^{-1}(x)$ must lie on the line $y = x$. Equating $f(x)$ to $x$:
$x^2 - 2x - 4 = x$

2. Solve the resulting quadratic equation and reject the invalid root.

$x^2 - 3x - 4 = 0$
$(x - 4)(x + 1) = 0$
$x = 4 \quad \text{or} \quad x = -1$
Since $x$ must be in the domain of $f$ ($x \ge 3.5$), we reject $x = -1$.
$x = 4$

1. Find the range of the inner function $g(x)$.

For $x \ge -1$ and $a > 0$, we have $ax \ge -a$.
Since the exponential function is strictly increasing, $e^{ax} \ge e^{-a} > 0$.
Thus, $g(x) = 4 + e^{ax} \ge 4 + e^{-a}$.
The range of $g$ is $R_g = [4 + e^{-a}, \infty)$.

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

For $fg$ to exist, we must show that $R_g \subseteq D_f$.
Since $e^{-a} > 0$ for any $a > 0$, the minimum value of $g(x)$ satisfies:
$4 + e^{-a} > 4 > 3.5$
Therefore, $R_g \subseteq [3.5, \infty)$.
Given that the domain of $f$ is $D_f = [3.5, \infty)$, we have $R_g \subseteq D_f$.
This satisfies the condition for the composite function $fg$ to exist.

3. Find the algebraic expression for $fg(x)$.

Substitute $g(x)$ into $f(x)$:
$fg(x) = f(4 + e^{ax})$
$= |4 + 2(4 + e^{ax}) - (4 + e^{ax})^2|$
$= |12 + 2e^{ax} - (16 + 8e^{ax} + e^{2ax})|$
$= |-4 - 6e^{ax} - e^{2ax}|$
Since $e^{ax} > 0$ for all real $x$, the expression inside the modulus ($-4 - 6e^{ax} - e^{2ax}$) is always strictly negative.
Thus, removing the modulus gives the negation of the expression:
$fg(x) = 4 + 6e^{ax} + e^{2ax}$

4. Determine the exact range of $fg$.

Let $y = e^{ax}$. The function $4 + 6y + y^2$ represents an upward-facing parabola with its vertex at $y = -3$.
Since $y = e^{ax} > 0$, the function is strictly increasing for all valid inputs in our domain.
From Step 1, $e^{ax} \ge e^{-a}$ for $x \ge -1$. The minimum value of $fg$ occurs when $x = -1$:
Minimum value $= 4 + 6e^{-a} + (e^{-a})^2 = 4 + 6e^{-a} + e^{-2a}$
As $x \to \infty$, $e^{ax} \to \infty \implies fg(x) \to \infty$.
Therefore, the exact range of $fg$ is $[4 + 6e^{-a} + e^{-2a}, \infty)$.

1. Analyze the sign of the numerator and deduce the condition for the denominator.

For the function $g(x) = 4 + e^{ax}$, since $e^{ax} > 0$ for all real $x$, the numerator $g(x) > 0$ for all $x$ in its domain.
Therefore, for the inequality $\frac{g(x)}{x^{2}-2x-2} \ge 0$ to hold, the denominator must be strictly positive:
$x^{2}-2x-2 > 0$

2. Solve the quadratic inequality for the denominator.

Find the roots of the quadratic equation $x^{2}-2x-2 = 0$ using the quadratic formula:
$x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$
Since the coefficient of $x^2$ is positive, the parabola opens upwards. Thus, $x^{2}-2x-2 > 0$ gives:
$x < 1-\sqrt{3} \quad \text{or} \quad x > 1+\sqrt{3}$

3. Intersect the solution with the domain of $g(x)$.

The inequality is only valid where the function $g(x)$ is defined. The domain of $g$ is given as $x \ge -1$.
We note that $\sqrt{3} \approx 1.732$, so $1-\sqrt{3} \approx -0.732$.
Since $-1 < 1-\sqrt{3}$, we intersect the domain condition $x \ge -1$ with the intervals $x < 1-\sqrt{3}$ or $x > 1+\sqrt{3}$ to obtain the final exact range:
$-1 \le x < 1-\sqrt{3} \quad \text{or} \quad x > 1+\sqrt{3}$