For part (a), you do not even need to be aware of the simplification mentioned by FUZxxl: you can take a purely mechanical approach.
Note first of all (this is kind of a joke, but not quite) that
$a_n=a_n$
Now to get $a_{n+1}$, we take the sum up to $i=n$, then add the next term, which is $(n=1)^2-2(n+1)+1$. It is convenient, just to save space, to notice that $j^2-2j+1=(j-1)^2$, so to get to $a_{n+1}$, we added $n^2$ to $a_n$. Thus $a_{n+1}=a_n+ n^2$
Similarly, to get $a_{n+2}$, we add $(n+1)^2$ to $a_{n+1}$. Thus $a_{n+2}=a_{n+1}+ (n+1)^2=a_n+n^2+(n+1)^2$
The same sort of reasoning shows that $a_{n+3}=a_n+n^2+(n+1)^2+(n+2)^2$ and that $a_{n+4}=a_n+n^2+(n+1)^2+(n+2)^2+(n+3)^2$
Now "plug in" the values we found into the expression $a_{n+4}-4a_{n+3}+6a_{n+2}-4a_{n+1}+a_n$
Simplify, using algebra. If you do the simplification cleverly, it will not even be a lot of work.
After some manipulation, you will get $0$. All this work for nothing (feeble pun).
The point I am making is that once you get a grasp of what $a_k$ actually means, the verification of the desired identity is purely mechanical.
Naturally there are better (or at least slicker) ways of doing the job. However, for such slicker ways one needs somewhat more theoretical background, which at this stage you may not yet have. Maybe you almost do. If you are beginning to be comfortable with generating functions ("series"), you can look for a generating function argument.
I will assume that you can do the other parts. If difficulties remain, please leave a comment, and I or others may be able to help.