曲线 $C$ 的参数方程为
$x = t^2 + 2, \quad y = t^3, \quad \text{where } t \in \mathbb{R}.$

1. Find the Cartesian equation of the curve to determine its domain and shape (or sketch directly using a graphing calculator).

$x = t^2 + 2 \implies t^2 = x - 2$
$y = t^3 \implies y^2 = (t^2)^3 = (x-2)^3$
Since $t \in \mathbb{R}$, $t^2 \ge 0 \implies x \ge 2$.
The curve is given by $y^2 = (x-2)^3$ defined for $x \ge 2$.

2. Identify key features and describe the sketch of the curve.

At $t = 0$, $x = 2$ and $y = 0$, so the $x$-intercept is at $(2, 0)$.
The gradient $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t$.
At $t = 0$, $\frac{dy}{dx} = 0$, meaning there is a horizontal tangent (a cusp) at $(2, 0)$.
The equation $y^2 = (x-2)^3$ shows symmetry about the $x$-axis. The curve is drawn in the first and fourth quadrants, curving outwards as $x \to \infty$.

1. Determine the value of the parameter $t$ corresponding to the point $(6, 8)$.

Substitute $y = 8$ into the equation for $y$:
$t^3 = 8 \implies t = 2$
Verify with the $x$-coordinate:
$x = (2)^2 + 2 = 6$
Since this is consistent, $t = 2$ at the point $(6, 8)$.

2. Differentiate the parametric equations with respect to $t$ to find $\frac{dy}{dx}$.

$\frac{dx}{dt} = 2t$
$\frac{dy}{dt} = 3t^2$
Using the chain rule:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{2t} = \frac{3t}{2}$

3. Calculate the gradient of the tangent at $t = 2$ and formulate its equation.

Substitute $t = 2$ into $\frac{dy}{dx}$:
$m = \frac{3(2)}{2} = 3$
Using the point-gradient form for the tangent equation at $(6, 8)$:
$y - 8 = 3(x - 6)$
$y - 8 = 3x - 18$
$y = 3x - 10$

1. Retrieve or derive the equation of the tangent to $C$ at $(6, 8)$ from the preceding part.

At the point $(6, 8)$, $t^3 = 8 \implies t = 2$.

From parametric differentiation:
$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$
$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t$

At $t = 2$, the gradient $m = \frac{3}{2}(2) = 3$.
The equation of the tangent is:
$y - 8 = 3(x - 6) \implies y = 3x - 10$

2. Substitute the parametric equations of $C$ into the tangent equation to find intersection points.

Substitute $x = t^2 + 2$ and $y = t^3$ into $y = 3x - 10$:
$t^3 = 3(t^2 + 2) - 10$
$t^3 - 3t^2 + 4 = 0$

3. Solve the polynomial equation for $t$.

Since the tangent touches the curve at $t = 2$, $t = 2$ must be a repeated root of this equation. By factorisation (or using a graphing calculator):
$(t - 2)^2(t + 1) = 0$
For the point $P$ where the line meets the curve again, $t + 1 = 0 \implies t = -1$.

4. Find the coordinates of $P$ using the parameter $t = -1$.

Substitute $t = -1$ into the parametric equations of $C$:
$x = (-1)^2 + 2 = 3$
$y = (-1)^3 = -1$

The coordinates of point $P$ are $(3, -1)$.

1. Find the gradient of the normal to the curve at the point where $t = m$.

Since $x = t^2 + 2$ and $y = t^3$, we have $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$.
The gradient of the tangent is $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t$.
At $t=m$, the gradient of the tangent is $\frac{3}{2}m$.
Therefore, the gradient of the normal is $-\frac{1}{\frac{3}{2}m} = -\frac{2}{3m}$.

2. Formulate the equation of the normal line.

Using the point-gradient form, the equation of the normal at $\left(m^2+2, m^3\right)$ is:
$y - m^3 = -\frac{2}{3m} \left(x - (m^2+2)\right)$
Multiplying by $3m$ to eliminate the fraction:
$3my - 3m^4 = -2x + 2m^2 + 4$
$2x + 3my = 3m^4 + 2m^2 + 4$

3. Find the coordinates of the $x$ and $y$ intercepts, $Q$ and $R$.

To find $Q$ (the $x$-intercept), set $y=0$:
$2x = 3m^4 + 2m^2 + 4 \implies x = \frac{3m^4 + 2m^2 + 4}{2}$
So, $Q$ is $\left(\frac{3m^4 + 2m^2 + 4}{2}, 0\right)$.

To find $R$ (the $y$-intercept), set $x=0$:
$3my = 3m^4 + 2m^2 + 4 \implies y = \frac{3m^4 + 2m^2 + 4}{3m}$
So, $R$ is $\left(0, \frac{3m^4 + 2m^2 + 4}{3m}\right)$.

4. Calculate the coordinates of the midpoint $F$.

Since $F$ is the midpoint of $QR$, its coordinates are the average of the coordinates of $Q$ and $R$:
$x_F = \frac{\frac{3m^4 + 2m^2 + 4}{2} + 0}{2} = \frac{3m^4 + 2m^2 + 4}{4}$
$y_F = \frac{0 + \frac{3m^4 + 2m^2 + 4}{3m}}{2} = \frac{3m^4 + 2m^2 + 4}{6m}$
Hence, the coordinates of $F$ are $\left( \frac{3m^4 + 2m^2 + 4}{4}, \frac{3m^4 + 2m^2 + 4}{6m} \right)$.