first note that you don't need to assume the group is abelian, because it follows from the fact that the order of each element (except identity) is 2.

Spoiler:

then, assume G has more than one element and consider subgroup A1 of G generated by a nonzero element. Order of A1 is 2. But G/A1 is also a group in which every element is an involution... if it has more than one element you can take a quotient again...and so on. This process must terminate because G is finite and every step decreases number of elemts to one half.

for 2) we have n>2 so n >= 4. select two distinct nonzero elements u,v and consider subgroup H = {0,u,v,u+v} (these are 4 distinct elements).

then, partition G into cosets by H. You can rearrange the summation a1+...+an such that you first add elements from each coset a+H and then add those results all together. But summation of elements of coset a+H gives you (a+a+a+a)+ (0+u+v+(u+v)) = 0.