The function $\mathrm{f}$ is given by
$\mathrm{f}(x) = \begin{cases} \frac{5}{2} - x, & x \in \mathbb{R},\ x < 2 \\ \frac{1}{x}, & x \in \mathbb{R},\ x \ge 2 \end{cases}$

1. Analyse the linear part of the function for $x < 2$ to find any intercepts.

For $x < 2$, $\mathrm{f}(x) = \frac{5}{2} - x$, which is a straight line with gradient $-1$.
When $x = 0$, $y = \frac{5}{2} - 0 = \frac{5}{2}$. The $y$-intercept is $(0, \frac{5}{2})$.
To find the $x$-intercept, set $y = 0 \implies x = \frac{5}{2}$. However, $\frac{5}{2} > 2$, so this point is outside the domain $x < 2$. Thus, this part of the graph does not cross the $x$-axis.
At the domain boundary $x = 2$, $\mathrm{f}(2)$ approaches $\frac{5}{2} - 2 = \frac{1}{2}$.

2. Analyse the rational part of the function for $x \ge 2$ to find asymptotes and continuation.

For $x \ge 2$, $\mathrm{f}(x) = \frac{1}{x}$.
At $x = 2$, $\mathrm{f}(2) = \frac{1}{2}$. Since this matches the boundary value of the linear part, the graph is continuous at $(2, \frac{1}{2})$.
As $x \to \infty$, $\frac{1}{x} \to 0$. Therefore, $y = 0$ is a horizontal asymptote.
Since $\frac{1}{x} > 0$ for all $x \ge 2$, this part does not cross the $x$-axis.

3. Synthesize the findings into the final sketch properties.

The combined sketch should show:

1. A straight line segment starting from the top left, passing through $(0, \frac{5}{2})$ and ending at $(2, \frac{1}{2})$.
2. A smooth curve starting from $(2, \frac{1}{2})$ decreasing gradually and approaching the horizontal line $y=0$.
3. Explicit labels for the asymptote $y=0$ and the intercept $(0, \frac{5}{2})$.

1. Explain the existence of the inverse function based on the graph from part (i).

From the graph of $y = \mathrm{f}(x)$ sketched in part (i), any horizontal line $y = k$ intersects the curve at most once. Therefore, $\mathrm{f}$ is a one-to-one function, which means its inverse $\mathrm{f}^{-1}$ exists.

2. Find the inverse rule and its domain for the first branch ($x < 2$).

Let $y = \frac{5}{2} - x$.
Making $x$ the subject gives $x = \frac{5}{2} - y$.
The domain for this piece is $x < 2$.
Substituting the expression for $x$, we get $\frac{5}{2} - y < 2 \implies y > \frac{1}{2}$.
Thus, one piece of the inverse is $\mathrm{f}^{-1}(x) = \frac{5}{2} - x$ for $x > \frac{1}{2}$.

3. Find the inverse rule and its domain for the second branch ($x \ge 2$).

Let $y = \frac{1}{x}$.
Making $x$ the subject gives $x = \frac{1}{y}$.
The domain for this piece is $x \ge 2$.
Substituting the expression for $x$, we get $\frac{1}{y} \ge 2$.
Since $x \ge 2$, $y$ must be positive. Multiplying by $y$ gives $1 \ge 2y \implies 0 < y \le \frac{1}{2}$.
Thus, the other piece of the inverse is $\mathrm{f}^{-1}(x) = \frac{1}{x}$ for $0 < x \le \frac{1}{2}$.

4. Combine the results into a single piecewise function in a similar form.

Writing the branches in ascending order of the domain of $x$:
$\mathrm{f}^{-1}(x) = \begin{cases} \frac{1}{x}, & x \in \mathbb{R},\ 0 < x \le \frac{1}{2} \\ \frac{5}{2} - x, & x \in \mathbb{R},\ x > \frac{1}{2} \end{cases}$

1. State the range of the inner function, $\mathrm{f}^{-1}$.

The range of an inverse function $\mathrm{f}^{-1}$ is identically the domain of the original function $\mathrm{f}$. From the piecewise definition in the stem, $\mathrm{f}(x)$ is defined for all real numbers.
$\mathrm{R}_{\mathrm{f}^{-1}} = \mathrm{D}_{\mathrm{f}} = \mathbb{R}$

2. Check the condition for the existence of the composite function.

For the composite function $\mathrm{gf}^{-1}$ to exist, the range of $\mathrm{f}^{-1}$ must be a subset of the domain of $\mathrm{g}$.
The domain of $\mathrm{g}$ is given as $\mathrm{D}_{\mathrm{g}} = [2, \infty)$.
Since $\mathbb{R} \not\subseteq [2, \infty)$, we have $\mathrm{R}_{\mathrm{f}^{-1}} \not\subseteq \mathrm{D}_{\mathrm{g}}$.
Therefore, $\mathrm{g}\mathrm{f}^{-1}$ does not exist.