I think I have solved the problem. I want to verify my proof, since I don't have a teacher to help me.
Proof: Since, ${360=8*45}$ and ${gcd(45,8)=1}$, hence if we can prove that ${45|a^2(a^2-1)(a^2-4)}$ and ${8|a^2(a^2-1)(a^2-4)}$ , then we are done proving the question.
Now, to prove divisibility by ${45}$, we see that ${45=9*5}$, also ${gcd(9,5)=1}$ and thus we can prove divisibility by ${45}$ if we can prove ${9|a^2(a^2-1)(a^2-4)}$ and ${5|a^2(a^2-1)(a^2-4)}$.
Since ${a}$ is an arbitrary integer, it may be represented as ${2k}$ or ${2k+1}$. So, ${a^2(a^2-1)(a^2-4)=a^2(a+1)(a-1)(a+2)(a-2)}$ may be represented as ${4k^2(2k+1)(2k-1)(2k+2)(2k-2)=16k^3(2K-1)(2k+1)(k+1)}$, which is divisible by ${8}$. Similarly, for ${a=2k+1}$, we prove the divisibility by substituting $a$ by ${2k+1}$ to get ${(2k+1)^2(2k+2)(2k)(2k+3)(2k-1)=4k(k+1)(2k+1)^2(2k-1)(2k+3)}$ which is divisible by ${8}$. Therefore ${8|a^2(a^2-1)(a^2-4)}$ . Similarly, to prove divisibility by $9$, we represent $a$ by ${3k}$, ${3k+1}$ and ${3k+2}$ and prove divisibility for all of them in a similar manner. Same method is used to prove divisibility by $5$ by represent $a$ by $5k,5k+1,5k+2,5k+3,5k+4$ and prove divisibility for all of these cases. (While doing in paper, I actually prove these divisibilities.) Hence, ${8|a^2(a^2-1)(a^2-4)}$ and ${45|a^2(a^2-1)(a^2-4)}$. Since, $gcd(8,45)=1$, so ${(8*45)|a^2(a^2-1)(a^2-4)}$ . Therefore, it is proved that ${360|a^2(a^2-1)(a^2-4)}$ .
Now, please tell me if my method is correct or not. If incorrect, then please show me how to solve it properly otherwise please show me how to improve my solution. Thank you! :)
Perhaps not as formal, but easy to follow if you look at it like this: $$360 = 2^3 \cdot 3^2 \cdot 5$$ and $$a^2(a^2-1)(a^2-4) = (a-2)(a-1)a^2(a+1)(a+2)$$ Notice that this is a product of five consecutive integers (with the middle one squared), so: