The parabola is g(x) = ax² + q.

Step 1: Analyze the given information and graph.
The parabola is $g(x) = ax^2 + q$.
The straight line is $f(x) = mx + c$.
R and S(2; 0) are the x-intercepts of $g$.
T(0; 8) is the y-intercept of $g$.
Graph $f$ passes through R and T.

Since $g(x) = ax^2 + q$ is a parabola with its axis of symmetry at $x=0$ (the y-axis), and S(2; 0) is an x-intercept, the other x-intercept R must be symmetric to S with respect to the y-axis. Therefore, R is (-2; 0).
T(0; 8) is the y-intercept of $g$. For a parabola of the form $ax^2+q$, the y-intercept is also the vertex. Since the parabola has x-intercepts and the vertex is above the x-axis, it opens downwards.

5.1: Write down the range of $g$.
The vertex of $g(x) = ax^2 + q$ is at $(0, q)$. From the graph, T(0, 8) is the vertex. Since the parabola opens downwards, the maximum value of $g(x)$ is the y-coordinate of the vertex.
The maximum value is $y=8$.
The range of $g$ is all real numbers less than or equal to 8.

$$\text{Range of } g: \boxed{y \le 8}$$

5.2: Write down the x-coordinate of R.
As established in Step 1, due to the symmetry of $g(x) = ax^2 + q$ about the y-axis, and S(2; 0) being an x-intercept, R must be the symmetric x-intercept.

$$\text{x-coordinate of R: } \boxed{-2}$$

5.3: Calculate the values of $a$ and $q$.
From T(0; 8) being the y-intercept of $g(x) = ax^2 + q$, we substitute $x=0$ and $g(x)=8$:

$$8 = a(0)^2 + q$$ $$8 = q$$

So, $q=8$. The equation for $g(x)$ is now $g(x) = ax^2 + 8$.
We use an x-intercept, for example S(2; 0), to find $a$. Substitute $x=2$ and $g(x)=0$:

$$0 = a(2)^2 + 8$$ $$0 = 4a + 8$$ $$4a = -8$$ $$a = \frac{-8}{4}$$ $$a = -2$$

The values are $a = -2$ and $q = 8$.

$$\boxed{a = -2, q = 8}$$

5.4: Determine the equation of $f$.
The function $f(x) = mx + c$ is a straight line that passes through R(-2; 0) and T(0; 8).
First, calculate the slope $m$:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{0 - (-2)} = \frac{8}{2} = 4$$

Next, find the y-intercept $c$. Since T(0; 8) is a point on the line and its x-coordinate is 0, the y-coordinate 8 is the y-intercept.
So, $c = 8$.
The equation of $f$ is $f(x) = 4x + 8$.

$$\boxed{f(x) = 4x + 8}$$

5.5: Use the graphs to determine the value(s) of $x$ for which:

5.5.1: $f(x) = g(x)$
This condition means finding the x-coordinates where the graphs of $f$ and $g$ intersect. From the graph, the intersection points are R and T.
R has x-coordinate -2.
T has x-coordinate 0.

$$\boxed{x = -2 \text{ or } x = 0}$$

5.5.2: $x \cdot g(x) \le 0$
We know $g(x) = -2x^2 + 8$. The x-intercepts of $g(x)$ are $x=-2$ and $x=2$.
The parabola $g(x)$ is above the x-axis (i.e., $g(x) \ge 0$) for $x \in [-2, 2]$.
The parabola $g(x)$ is below the x-axis (i.e., $g(x) \le 0$) for $x \in (-\infty, -2] \cup [2, \infty)$.

We need $x \cdot g(x) \le 0$. This occurs when:

1. $x \le 0$ AND $g(x) \ge 0$: This means $x \in (-\infty, 0]$ and $x \in [-2, 2]$. The intersection is $x \in [-2, 0]$.
2. $x \ge 0$ AND $g(x) \le 0$: This means $x \in [0, \infty)$ and $x \in (-\infty, -2] \cup [2, \infty)$. The intersection is $x \in [2, \infty)$.

Combining these two cases, the solution is $x \in [-2, 0] \cup [2, \infty)$.

$$\boxed{x \in [-2, 0] \cup [2, \infty)}$$

5.6: The graph $h$ is obtained when $g$ is reflected along the line $y=0$. Write down the equation of $h$ in the form $h(x) = px^2 + k$.
Reflecting a function $y = g(x)$ along the line $y=0$ (the x-axis) means that the new y-values are the negative of the original y-values. So, $h(x) = -g(x)$.
We found $g(x) = -2x^2 + 8$.

$$h(x) = -(-2x^2 + 8)$$ $$h(x) = 2x^2 - 8$$

This is in the form $h(x) = px^2 + k$, where $p=2$ and $k=-8$.

$$\boxed{h(x) = 2x^2 - 8}$$