The function $\text{f}$ is defined by
$\text{f} : x \mapsto (x-2)^2 - 4,\quad x \in \mathbb{R},\ x \ge 0.$

1. Identify key coordinates (vertex, intercepts) to sketch the graph of $\text{f}$.

The function $\text{f}(x) = (x-2)^2 - 4$ is a parabola opening upwards.
The vertex is at $(2, -4)$.
Find the $y$-intercept by substituting $x=0$:
$\text{f}(0) = (0-2)^2 - 4 = 0$.
Find the $x$-intercepts by setting $\text{f}(x) = 0$:
$(x-2)^2 - 4 = 0 \implies x-2 = \pm 2 \implies x = 0 \text{ or } x = 4$.
Since the domain is $x \ge 0$, the sketch must show a solid curve starting at the origin $(0,0)$, going down to the minimum point at $(2,-4)$, and curving upwards to pass through $(4,0)$.

2. Explain why the inverse function $\text{f}^{-1}$ does not exist using the concept of one-to-one mapping.

Any horizontal line drawn between $y=0$ and $y=-4$ (exclusive) intersects the graph of $\text{f}$ at two points.
For example, $\text{f}(0) = 0$ and $\text{f}(4) = 0$.
Since two distinct elements in the domain map to the same value in the codomain, $\text{f}$ is not a one-to-one function. Therefore, its inverse $\text{f}^{-1}$ does not exist.

1. Identify the smallest $k$ that makes the function one-to-one on $x \ge k$

For $\text{f}^{-1}$ to exist, $\text{f}$ must be a one-to-one function.
The turning point of $\text{f}(x) = (x-2)^2 - 4$ is at $x = 2$.
For the domain $x \ge k$, the function is strictly increasing (and thus one-to-one) if $k \ge 2$.
Therefore, the smallest value of $k$ is $2$.

1. Determine the domain of the inverse function by finding the range of the restricted original function.

From part (ii), the smallest value for $k$ is $2$, so the restricted domain of $\text{f}$ is $x \ge 2$.
For $x \ge 2$, the minimum point of $\text{f}(x) = (x-2)^2 - 4$ is at $(2, -4)$.
Therefore, the range of $\text{f}$ is $\text{f}(x) \ge -4$.
Since the domain of $\text{f}^{-1}$ is the range of $\text{f}$, the domain of $\text{f}^{-1}$ is $x \ge -4$.

2. Set $y = \text{f}(x)$ and begin isolating $x$.

Let $y = (x-2)^2 - 4$.
$(x-2)^2 = y + 4$
$x - 2 = \pm\sqrt{y+4}$

3. Use the restricted domain to choose the correct sign and write the function in the required form.

Since the domain of $\text{f}$ is $x \ge 2$, we have $x - 2 \ge 0$.
Therefore, we must reject the negative square root and take the positive square root:
$x - 2 = \sqrt{y+4}$
$x = 2 + \sqrt{y+4}$
Expressing this in the required mapping form with its domain:
$\text{f}^{-1} : x \mapsto 2 + \sqrt{x+4}, \quad x \in \mathbb{R}, \ x \ge -4$

1. Find the expression for $\text{f}^{-1}(x)$ based on the domain restriction from parts (ii) and (iii).

Let $y = \text{f}(x) = (x-2)^2 - 4$.
$(x-2)^2 = y + 4$
Since the domain is restricted to $x \ge 2$ for the inverse to exist, we take the positive square root:
$x - 2 = \sqrt{y+4} \implies x = 2 + \sqrt{y+4}$
Thus, $\text{f}^{-1}(x) = 2 + \sqrt{x+4}$.

2. Substitute $\text{g}(x)$ into $\text{f}^{-1}(x)$ to find the expression for the composite function.

$\text{f}^{-1}\text{g}(x) = \text{f}^{-1}(3x-5)$
$\text{f}^{-1}\text{g}(x) = 2 + \sqrt{(3x-5) + 4}$
$\text{f}^{-1}\text{g}(x) = 2 + \sqrt{3x-1}$

3. Determine the range of $\text{f}^{-1}\text{g}$ using the domain of $\text{g}$.

The domain of $\text{g}$ is $x \ge 3$.
For $x \ge 3$:
$3x - 1 \ge 3(3) - 1 = 8$
Therefore, $\sqrt{3x-1} \ge \sqrt{8} = 2\sqrt{2}$.
Hence, $\text{f}^{-1}\text{g}(x) = 2 + \sqrt{3x-1} \ge 2 + 2\sqrt{2}$.
The range is $[2+2\sqrt{2}, \infty)$.