We observe \[ a_n = (n^2-3)(n^2+2)(n^2+6). \]With this, let $p$ be any prime. Then since at least one of $(3/p)$, $(-2/p)$ and $(-6/p)$ is 1 (notice that if all three are -1, then $(-6/p)=(-2/p)(3/p)=1$, a contradiction). This proves (a). As for (b), suppose $N$ is such that $v_2(N)=31$ (the number 31 is arbitrary). I show none of $a_n$ is divisible by $N$. Suppose $N\mid a_n$. If $n$ is odd, then, $2^{31}\mid n^2-3\equiv 2\pmod{4}$, a contradiction. Likewise, if $n$ is even, then $v_2(n^2+2)=v_2(n^2+6)=1$, so $v_2(a_n)=2$, again a contradiction. We now tackle (c), noticing $1989 = 3^2\cdot 13\cdot 17$. Start with mod 9. If $3\mid n$, then $v_3(a_n) = 2$. If $3\nmid n$, then $n^2\equiv 7\pmod{9}$ must hold, so $n\in\{4,5\}\pmod{9}$. For modulo 17, notice that it is congruent to 2 modulo 3 and 1 modulo 8. So, $(3/17)=-1$, $(-2/17)=1$ and $(-6/17)=-1$. So, $17\mid a_n$ iff $17\mid n^2+2$, so $n\in\{\pm 7\}\pmod{17}$. Lastly, for $13$ we have $(-2/13)=(2/13)=-1$ since $13\equiv 5\pmod{8}$, so $13\mid n^2-3$. Thus, $n\in\{\pm 4\}\pmod{13}$. These uniquely recover allowed $n$ in modulo 1989, the rest is boring calculations that I omit.