The functions f and g are defined by $f: x \mapsto |(x+1)(x-3)|,\quad x \in \mathbb{R},$ $g: x \mapsto x-2,\quad x \in \mathbb{R}\text{ and }x \ge -1.$

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

For $f(x) = |(x+1)(x-3)|$, the modulus ensures that $f(x) \ge 0$ for all $x \in \mathbb{R}$. Since $f(3) = 0$, the minimum value is $0$. As $x \to \infty$, $f(x) \to \infty$.
Thus, the range of $f$ is $R_f = [0, \infty)$.

2. Compare the range of $f$ with the domain of $g$ to show that $gf$ exists.

The domain of $g$ is given as $x \ge -1$, so $D_g = [-1, \infty)$.
Since $R_f = [0, \infty)$ and $[0, \infty) \subseteq [-1, \infty)$, we have $R_f \subseteq D_g$.
Therefore, the composite function $gf$ exists.

3. Determine the range of the composite function $gf$.

The domain of $g$ when evaluating $gf(x)$ is restricted to the range of $f$, which is $[0, \infty)$.
Evaluating $g(y) = y - 2$ for $y \in [0, \infty)$ yields $g(y) \ge 0 - 2 = -2$.
Hence, the range of $gf$ is $[-2, \infty)$.

1. Show that $f$ is not one-one by providing a counterexample where two distinct inputs map to the same output.

Testing the roots of the quadratic inside the modulus:
$f(-1) = |(-1+1)(-1-3)| = 0$
$f(3) = |(3+1)(3-3)| = 0$

Since $f(-1) = f(3) = 0$ but $-1 \neq 3$, there are multiple $x$-values that map to the same value of $f(x)$. Thus, $f$ is not one-one.

2. Analyse the behaviour and turning points of $f(x)$ to find the largest $k$ for the restricted domain $x \le k$.

The inner expression $y = (x+1)(x-3)$ is a parabola opening upwards with $x$-intercepts at $x = -1$ and $x = 3$. Its line of symmetry and minimum point is at $x = 1$.
Taking the modulus, $f(x) = |(x+1)(x-3)|$ will reflect the negative portion (from $x=-1$ to $x=3$) above the $x$-axis.
From $x = -\infty$ to $x = -1$, the graph is strictly decreasing towards $0$.
From $x = -1$ to $x = 1$, the graph is strictly increasing towards a local maximum.
For the function to be one-one on a domain $x \le k$, it must remain strictly decreasing without changing direction. Therefore, the domain cannot extend past the first turning point at $x = -1$.
Hence, the largest possible value of $k$ is $-1$.

1. Determine the expression for $f(x)$ for the restricted domain and state the domain of $f^{-1}$.

From part (ii), the restricted domain for $f$ to be one-one is $x \le -1$.
For $x \le -1$, the expression $(x+1)(x-3) \ge 0$, so we can remove the modulus sign:
$f(x) = x^2 - 2x - 3$

The range of $f$ on this restricted domain is $f(x) \ge 0$.
Since the domain of $f^{-1}$ is the range of $f$, the domain of $f^{-1}$ is $x \ge 0$.

2. Set $y = f(x)$ and make $x$ the subject by completing the square.

Let $y = f(x)$.
$y = x^2 - 2x - 3$
Completing the square for the quadratic expression:
$y = (x - 1)^2 - 1^2 - 3$
$y = (x - 1)^2 - 4$
$(x - 1)^2 = y + 4$
$x - 1 = \pm\sqrt{y + 4}$

3. Select the correct root based on the domain restriction and state $f^{-1}$ in the required form.

Since the domain is restricted to $x \le -1$, it follows that $x - 1 \le -2 < 0$.
Therefore, we must reject the positive root and take the negative root:
$x - 1 = -\sqrt{y + 4}$
$x = 1 - \sqrt{y + 4}$

Expressing the inverse function in a similar form to the original functions in the question:
$f^{-1}: x \mapsto 1 - \sqrt{x+4}, \quad x \ge 0$