Not hard to show that $a_n=\binom{2n}{n}$ for all $n\in \mathbb{Z}^+$. The rest is easy.

In fact, part (b) with $a_n = \tbinom{2n}{n}$ is integral in Erdos' proof of Bertrand's postulate

For (c) we can use Wolstenholme's theorem.

Borbas wrote: For (c) we can use Wolstenholme's theorem. It's a much easier special case.

You can actually indeed use Wolstenholme to prove a stronger result for (c), namely $p^3 \mid a_p-2$. For that one has to show a fact strongly related to Wolstenholme: \[ \binom{2p-1}{p-1} \equiv 1 \pmod {p^3} \quad \text{for all primes } p>3.\]The aforementioned result follows immediately using this. On a different regard: Suppose you are sloppy (like me) and do not see the binomial coefficient. Then one could have a super long solution to (a) as this: Step 1: By induction $a_n = \frac{2^n \cdot (2n-1)!!}{n!}$. Step 2: Modify the well-known Legendre formula $v_p(n!) = \sum_{i=1}^{\infty} \left \lfloor \frac{n}{p^i} \right \rfloor$ to \[ v_p((2n-1)!!) = \sum_{i=1}^{\infty} \left \lfloor \frac{2n-1}{p^i} \right \rfloor - \sum_{i=1}^{\infty} \left \lfloor \frac{2n-1}{2p^i} \right \rfloor. \]Step 3: By some bounding for primes $p>2$ we have $\left \lfloor \frac{4n-1}{2p^i} \right \rfloor = \left \lfloor \frac{4n-2}{2p^i} \right \rfloor$. Now it suffices to prove $v_p(2^n \cdot (2n-1)!!) \geq v_p(n!)$ for all primes $p$. It is easy for $p=2$. So we have reduced it to $v_p((2n-1)!!) \geq v_p(n!)$ for $p>2$. By Step 2 we must show \[ \sum_{i=1}^{\infty} \left \lfloor \frac{2n-1}{p^i} \right \rfloor - \sum_{i=1}^{\infty} \left \lfloor \frac{2n-1}{2p^i} \right \rfloor \geq \sum_{i=1}^{\infty} \left \lfloor \frac{n}{p^i} \right \rfloor. \qquad (*) \]But well, we have \[ \left\lfloor \frac{n}{p^i} \right \rfloor + \left\lfloor \frac{2n-1}{2p^i} \right \rfloor \leq \left\lfloor \frac{4n-1}{2p^i} \right \rfloor = \left\lfloor \frac{4n-2}{2p^i} \right\rfloor = \left\lfloor \frac{2n-1}{p^i} \right \rfloor, \]so \[ \left\lfloor \frac{2n-1}{p^i} \right \rfloor - \left\lfloor \frac{2n-1}{2p^i} \right \rfloor \geq \left\lfloor \frac{n}{p^i} \right \rfloor \]with which $(*)$ immediately follows. Can I say that it is painful to type floors with additional \left and \right's?

Here is a quick solution for part (c): $\binom{2n}{n}-2=\sum_{k=1}^{n-1}\binom{n}{k}^2\equiv 0\pmod{n^2}$