The functions f and g are defined by $f : x \mapsto 1-\dfrac{1}{x^{2}-1}$ for $x \in \mathbb{R},\; x<-1$, $g : x \mapsto x^{2}+4x+5$ for $x \in \mathbb{R},\; x\le 1$.

1. Set $y = f(x)$ and rearrange to solve for $x^2$.

Let $y = 1-\frac{1}{x^{2}-1}$

$\frac{1}{x^{2}-1} = 1 - y$

$x^{2}-1 = \frac{1}{1-y}$

$x^{2} = 1 + \frac{1}{1-y} = \frac{2-y}{1-y}$

2. Take the square root and choose the correct sign based on the domain of $f$.

$x = \pm \sqrt{\frac{2-y}{1-y}}$

Since the domain of $f$ is $x < -1$, $x$ must be negative, so we reject the positive root.
$x = -\sqrt{\frac{2-y}{1-y}}$

$\therefore f^{-1}(x) = -\sqrt{\frac{2-x}{1-x}}$

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

For $x < -1$, we have $x^2 > 1 \implies x^2 - 1 > 0$.
Since $\frac{1}{x^2-1} > 0$, it follows that $1 - \frac{1}{x^2-1} < 1$.

As $x \to -1^-$, $\frac{1}{x^2-1} \to \infty \implies f(x) \to -\infty$.
Thus, the range of $f$ is $y < 1$.

The domain of $f^{-1}$ is the range of $f$, which is $x < 1$.

1. Determine the ranges of functions $f$ and $g$.

\text{For } f(x) = 1 - \frac{1}{x^2-1} \text{ with } x < -1: \
x^2 > 1 \implies x^2 - 1 > 0 \implies \frac{1}{x^2-1} > 0 \
\therefore f(x) < 1 \
\text{As } x \to -1^-, f(x) \to -\infty \implies R_f = (-\infty, 1) \
\text{For } g(x) = x^2 + 4x + 5 = (x+2)^2 + 1 \text{ with } x \le 1: \
\text{The minimum is at } x = -2, \text{ where } g(-2) = 1. \
\text{As } x \to -\infty, g(x) \to \infty \implies R_g = [1, \infty)

2. Assess the existence of the composite function $fg$.

\text{For } fg \text{ to exist, we require } R_g \subseteq D_f. \
D_f = (-\infty, -1) \
\text{Since } R_g = [1, \infty) \not\subseteq (-\infty, -1), \
fg \text{ does not exist.}

3. Assess the existence of the composite function $gf$.

\text{For } gf \text{ to exist, we require } R_f \subseteq D_g. \
D_g = (-\infty, 1] \
\text{Since } R_f = (-\infty, 1) \subseteq (-\infty, 1], \
gf \text{ exists.}

4. Calculate the range of the existing composite function $gf$.

\text{The range of } gf \text{ is the range of } g \text{ evaluated over the domain } R_f = (-\infty, 1). \
\text{We consider } g(x) = (x+2)^2 + 1 \text{ for } x \in (-\infty, 1). \
\text{The turning point } x = -2 \text{ lies within this interval.} \
\text{Minimum value} = g(-2) = 1 \
\text{As } x \to -\infty, g(x) \to \infty \
\therefore R_{gf} = [1, \infty)