The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \cos\left(\frac{\pi x}{2}\right)$, $x \in \mathbb{R}$, $-2 < x \le 0$.

1. Find the coordinates of the end points.

Substitute $x = 0$:
$\mathrm{f}(0) = \cos(0) = 1$
Since $x=0$ is included in the domain ($-2 < x \le 0$), there is a solid dot at $(0, 1)$.

Substitute $x = -2$:
$\mathrm{f}(-2) = \cos(-\pi) = -1$
Since $x=-2$ is excluded from the domain, there is an open circle at $(-2, -1)$.

2. Find the coordinates of the axial intercepts.

The $y$-intercept occurs at $x=0$, which we already found to be $(0, 1)$.

For the $x$-intercept, set $y = 0$:
$\cos\left(\frac{\pi x}{2}\right) = 0$
Since $-2 < x \le 0$, the range for the angle is $-\pi < \frac{\pi x}{2} \le 0$.
$\frac{\pi x}{2} = -\frac{\pi}{2} \implies x = -1$
The $x$-intercept is $(-1, 0)$.

3. Describe the sketch of the graph.

Sketch a smooth curve representing half a period of a cosine wave (from a trough to a crest). The curve strictly increases from the open circle at $(-2, -1)$, passes through $(-1, 0)$, and terminates at the solid dot at $(0, 1)$.

1. Show that the inverse function exists using the horizontal line test.

Based on the graph of $y = \mathrm{f}(x)$ from part (i), any horizontal line $y = k$ intersects the curve at most once.
Since $\mathrm{f}$ is a one-to-one function, $\mathrm{f}^{-1}$ exists.

2. Let $y = \mathrm{f}(x)$ and rewrite the argument to fall within the principal range of inverse cosine.

Let $y = \cos\left(\frac{\pi x}{2}\right)$.
Since $\cos(\theta) = \cos(-\theta)$, we can equivalently write:
$y = \cos\left(-\frac{\pi x}{2}\right)$
Given the domain $-2 < x \le 0$, we multiply by $-\frac{\pi}{2}$ to get:
$0 \le -\frac{\pi x}{2} < \pi$
This interval lies entirely within the principal range $[0, \pi]$ of the inverse cosine function.

3. Solve for $x$ in terms of $y$ to determine the rule for the inverse function.

Taking the inverse cosine on both sides:
$-\frac{\pi x}{2} = \cos^{-1}(y)$
$x = -\frac{2}{\pi} \cos^{-1}(y)$
Therefore, the rule is $\mathrm{f}^{-1}(x) = -\frac{2}{\pi} \cos^{-1}(x)$.

4. Determine the domain of the inverse function.

The domain of $\mathrm{f}^{-1}$ is equal to the range of $\mathrm{f}$.
From the endpoints of $\mathrm{f}$ over $-2 < x \le 0$:
As $x \to -2$, $\mathrm{f}(x) \to \cos(-\pi) = -1$.
At $x = 0$, $\mathrm{f}(0) = \cos(0) = 1$.
Thus, the range of $\mathrm{f}$ is $(-1, 1]$.
$\mathrm{D}_{\mathrm{f}^{-1}} = (-1, 1]$

1. Determine the valid domain over which the inequality can be solved.

The inequality $\mathrm{g}(x) \ge \mathrm{f}(x)$ can only be evaluated where both functions are defined.
Domain of $\mathrm{f}$: $-2 < x \le 0$
Domain of $\mathrm{g}$: $-2 < x \le 2$

Intersecting these gives the common domain:
$x \in (-2, 0]$

2. Find the points of intersection of the two graphs using a graphing calculator.

Using a graphing calculator, plot $y_1 = (2x+1)^{\frac{2}{3}}$ and $y_2 = \cos\left(\frac{\pi x}{2}\right)$.
Solve for $y_1 = y_2$ within $-2 < x \le 0$:
$x \approx -0.67262 \quad \text{and} \quad x = 0$

3. Determine the required set of values from the graphs.

Observe the graphs to find where $y_1$ is above or touches $y_2$.
For $x \in (-2, -0.67262]$, $y_1 \ge y_2$.
For $x \in (-0.67262, 0)$, $y_1 < y_2$.
At exactly $x = 0$, $y_1 = y_2 = 1$, which satisfies $\mathrm{g}(x) \ge \mathrm{f}(x)$.

Thus, the set of values is:
$-2 < x \le -0.673 \text{ (to 3 s.f.) or } x = 0$

1. Find the range of $\mathrm{f}$ and verify the condition for the existence of $\mathrm{gf}$.

Given $\mathrm{f}(x) = \cos\left(\frac{\pi x}{2}\right)$ for $-2 < x \le 0$, the values of $\frac{\pi x}{2}$ lie in $(-\pi, 0]$.
The range of $\mathrm{f}$ is therefore $\mathrm{R}_{\mathrm{f}} = (-1, 1]$.
The domain of $\mathrm{g}$ is $\mathrm{D}_{\mathrm{g}} = (-2, 2]$.
Since $\mathrm{R}_{\mathrm{f}} \subseteq \mathrm{D}_{\mathrm{g}}$ (i.e. $(-1, 1] \subseteq (-2, 2]$), the composite function $\mathrm{gf}$ exists.

2. Determine the interval of values for the inner expression of $\mathrm{g}$.

To find the range of $\mathrm{gf}$, we evaluate $\mathrm{g}(y) = (2y+1)^{\frac{2}{3}}$ for $y \in \mathrm{R}_{\mathrm{f}}$, which means $y \in (-1, 1]$.
For $-1 < y \le 1$, we have:
$-2 < 2y \le 2$
$-1 < 2y+1 \le 3$

3. Find the range of the outer function over the calculated interval.

Let $u = 2y+1$. We need the range of $z = u^{\frac{2}{3}} = (u^{\frac{1}{3}})^2$ for $u \in (-1, 3]$.
Since the square of a real number is non-negative, the minimum value is $0$, occurring when $u=0$ (which is within the interval as it corresponds to $y = -0.5$).
The maximum value occurs at the upper boundary $u=3$, yielding $3^{\frac{2}{3}}$.
Thus, the range of $\mathrm{gf}$ is $[0, 3^{\frac{2}{3}}]$.