There are actually several different concepts here, so I'll try to address all of them. I'll get to the modular arithmetic in just a moment, but first a review:
SQUARE ROOTS
We should know that 25 has two square roots in ordinary arithmetic: 5 and -5.
MODULAR ARITHMETIC SQUARE ROOTS
IF the square root exists, there are 2 of them modulo a prime. To continue our example, 25 has the two square roots 5 and -5.
We can check this:
$$(-5)^2 = 25 \equiv 3\bmod 11$$
$$(5)^2 = 25 \equiv 3\bmod 11$$
To find the square roots sometimes takes a bit of trial and error. Often you have to go through each value $v$ and square it (to get $v^2$) to check if it's equivalent to $n \bmod p$, where $n$ is the number whose square root you want to find.
MULTIPLE PRIMES
Again, if a square root exists, there are two square roots modulo each prime. So if we are using multiple primes, there can be more square roots. For example, with two primes, there are 2 square roots modulo the first prime and two square roots modulo the second prime. This gives us $2 \cdot 2 = 4$ square roots.
In general, if we can find a square root modulo each prime, there are a total of $2^n$ square roots modulo $n$ primes.
RETURNING TO YOUR EXAMPLE
We can first calculate 3 modulo 11 and 13:
$$3 \equiv 3 (\bmod 11)$$
$$3 \equiv 3 (\bmod 13)$$
So, modulo 11, we are looking to find a number that, when squared, is equivalent to 3. If we find one, we know that there will be another. So we check the numbers: $1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 9, \dots$ and find that
$$5^2 = 25 \equiv 3 (\bmod 11)$$
...so we know that there will also be another square root modulo 11. Continuing on our quest, we check
$$6^2 = 36 \equiv 3 (\bmod 11)$$
...so we've found the square roots modulo 11. We then continue this modulo 13 to find that:
$$4^2 = 16 \equiv 3 (\bmod 13)$$
$$9^2 = 81 \equiv 3 (\bmod 13)$$
So we know that our square root is either 5 or 6 modulo 11, and either 4 or 9 modulo 13. This gives us 4 possibilities. We can then find that:
$$82 \equiv 5 (\bmod 11), 82 \equiv 4 (\bmod 13)$$
$$126 \equiv 5 (\bmod 11), 126 \equiv 9 (\bmod 13)$$
$$17 \equiv 6 (\bmod 11), 17 \equiv 4 (\bmod 13)$$
$$61 \equiv 6 (\bmod 11), 61 \equiv 9 (\bmod 13)$$
Best Answer
I'm used to seeing induction applied in a less 'systematic' way than you did, but let me try to shed some light on it for you. Mathematical induction is like a game of dropping dominoes. If you can guarantee that
then all pieces will fall. The first one is the basis step, wich you've already done. The second one is the induction hypoteses and the induction step. The induction hypoteses is equivalent to "If one piece falls", you're assuming that your formula is valid for an arbitrary natural number $k$. So that's a given. The induction step is equivalent to "then the next one will fall too", so using the given fact that it works for $k$, it must work for $k+1$ too.
Now to your case. The induction hypoteses gives us that $a_k = 5a_{k-1} + 8$ is congruent to three modulo 4, so $a_k \equiv3\;(\textrm{mod}\; 4)$. Now we need to evaluate if it is true for $a_{k+1}$. We need: $$a_{k+1} \equiv3\;(\textrm{mod}\; 4)$$ But we have: $$a_{k+1} = 5a_k + 8$$ And: $$8 \equiv 0 \;(\textrm{mod}\; 4)$$ $$5 \equiv 1 \;(\textrm{mod}\; 4)$$ Then: $$a_{k+1} = 5a_k + 8 \equiv 1a_k + 0 \equiv a_k \equiv 3\;(\textrm{mod}\; 4)$$ And now we end the proof by stating: "By the Principle of Mathematical Induction, the statement is valid for all $n \geq 0$". Q.E.D