Complete representative set of squares modulo $15$. 
$2.\,\,$Do the following computations.
$\text{(a)}$ Solve the equation $x^2\equiv 1\mod15$
Solution: We only need to choose a complete representative set modulo $15$ and verify the equation over such a set. In the following table, we choose the representative set 
  $$\{0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7\}$$ 
  and verify the equation as follows: $$
\begin{array}{c|c}
x   & 0 & \pm 1 & \pm 2 & \pm 3 & \pm 4 & \pm 5 & \pm 6 & \pm 7 \\
\hline
x^2 & 0 &    1  &   4   &  -6   &   1   &   -5  &   6   &    4   \\
\end{array}
$$
We see that the equation has four solutions: $\pm1$ and $\pm 4$.
(Note that $15$ is not a  prime, so we do not just have two square roots!)

I understand that if you do $8^2 \mod 15 $, $9^2  \mod 15$, $10^2 \mod 15$, and so forth, you get repetitions of the above representative set.  What I don't understand is how you could know this before doing those calculations manually.   

The set $\{1,2,3,...n-1\}$ is called complete residue set for number $n$. And all conguence relation modulo $n$ can be reduced to elements of this set.
Now for your question you can write $24$ as $(15 + 9)$ so $24^2 \equiv (15 + 9)^2 \equiv 15^2 + 270 + 9^2 \equiv 9^2 \pmod {15}$
And you need to check only the first $\frac{n}{2}$ numbers because we have: $9\equiv-6 \pmod{15}$ Now square and you'll get $(-6)^2 \equiv 6^2 \equiv 36 \pmod {15}$$
