Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle. Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle.

My work
We can describe the circle using polar coordinates:
$x(\theta) = \rho\cos(\theta)$, $y(\theta) = \rho\sin(\theta).$
$(x-8)^2 + y^2 = 16$ is a circle with radius of $4$. 
$\therefore \rho = 4$
Let $t = \theta \ \forall \ t \in [0, 2\pi]$ be our parameter.
$\therefore (x(t), y(t)) = (4\cos(t), 4\sin(t))$
$(x(t), y(t)) = (4\cos(\pi-t), 4\sin(\pi-t))$ represents the circle of radius $4$ and starting at $\pi$ and moving clockwise.
However, this circle is still centred at $(0, 0)$. To centre the circle at $(8,0)$, we must have the parameterisation $(x(t), y(t)) = (8 + 4\cos(\pi - t), 4\sin(\pi-t)) \ \forall \ t \in [0, 2\pi]$.
We can test values some values to get an indication of any errors in our parameterisation.
$(x(0), y(0)) = (4, 0)$
$(x(\pi/2), y(\pi/2)) = (8, 4)$
$(x(\pi), y(\pi)) = (12, 0)$
$(x(3\pi/2), y(3\pi/2)) = (8, -4)$
$(x(2\pi), y(2\pi)) = (4, 0)$

I would greatly appreciate it if people could please take the time to review my reasoning and solution for correctness.

Yes, the work is correct.
Below is a parametric plot of $\left\{8 + 4 \cos \left( \pi - t \right), 4 \sin \left(t\right)\right\}$ for $0\le t < 2\pi$.

To show that your parameterization starts at $(4,0)$ and proceeds in a clockwise manner, a sequence of values was created:
$$
\begin{array}{cccc}
 k & t & x(t) & y(t) \\\hline
 1 & 0 & 4 & 0 \\
 2 & \frac{\pi }{6} & 8-2 \sqrt{3} & 2 \\
 3 & \frac{\pi }{3} & 6 & 2 \sqrt{3} \\
 4 & \frac{\pi }{2} & 8 & 4 \\
 5 & \frac{2 \pi }{3} & 10 & 2 \sqrt{3} \\
 6 & \frac{5 \pi }{6} & 8+2 \sqrt{3} & 2 \\
 7 & \pi  & 12 & 0 \\
 8 & \frac{7 \pi }{6} & 8+2 \sqrt{3} & -2 \\
 9 & \frac{4 \pi }{3} & 10 & -2 \sqrt{3} \\
 10 & \frac{3 \pi }{2} & 8 & -4 \\
 11 & \frac{5 \pi }{3} & 6 & -2 \sqrt{3} \\
 12 & \frac{11 \pi }{6} & 8-2 \sqrt{3} & -2 \\
 1 & 0 & 4 & 0 \\
 2 & \frac{\pi }{6} & 8-2 \sqrt{3} & 2 \\
 3 & \frac{\pi }{3} & 6 & 2 \sqrt{3} \\
 4 & \frac{\pi }{2} & 8 & 4 \\
 5 & \frac{2 \pi }{3} & 10 & 2 \sqrt{3} \\
 6 & \frac{5 \pi }{6} & 8+2 \sqrt{3} & 2 \\
 7 & \pi  & 12 & 0 \\
 8 & \frac{7 \pi }{6} & 8+2 \sqrt{3} & -2 \\
 9 & \frac{4 \pi }{3} & 10 & -2 \sqrt{3} \\
 10 & \frac{3 \pi }{2} & 8 & -4 \\
 11 & \frac{5 \pi }{3} & 6 & -2 \sqrt{3} \\
 12 & \frac{11 \pi }{6} & 8-2 \sqrt{3} & -2 \\
\end{array}
$$
The $k$ values are plotted against the circle.
