How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution? 
Any help would be appreciated!

The numbers are small, so general techniques are not necessary. However, we describe, in this particular case, the method of Hensel Lifting. 
It is clear that the solutions modulo $7$ are $x\equiv \pm 3\pmod{7}$. We lift the solution $x\equiv 3\pmod{7}$ to a solution of $x^2\equiv 2\pmod{7^2}$. 
Any solution to $x^2\equiv 2\pmod{7^2}$ that is congruent to $3$ modulo $7$ has shape $x=3+7t$. Square. Modulo $7^2$, the result is congruent to $9+(2)(3)(7)t\pmod{7^2}$. We want this to be congruent to $2$ modulo $7^2$, so we want
$$9+(2)(3)(7)t\equiv 2\pmod{7^2}.$$
A little manipulation turns this to
$$(2)(3)t\equiv -1\pmod{7}.$$
Thus $t=1$ works, and we have
$$x\equiv 10\pmod{7^2}.$$
Now lift this solution to a solution of $x^2\equiv 2\pmod{7^3}$. We look for a solution of the shape $x\equiv 10+7^2t\pmod{7^3}$. 
Squaring, we get
$$100+(2)(10)(7^2)t\equiv 2\pmod{7^3},$$
which simplifies to 
$$(2)(10)t\equiv -2\pmod{7}.$$
A solution is $t\equiv 2\pmod{7}$. That gives solution $x\equiv 10+2(49)\pmod{7^3}$.
By general theory, the only solutions are therefore $x\equiv \pm 108\pmod{7^2}$.
Remark: If we wished to, we could continue, and lift to a solution modulo $7^4$, $7^5$, and so on. 
Hensel lifting is an important general technique for solving polynomial congruences modulo prime powers.  
