Further attempt for number 2:
suppose d=gcd(a+b,a-b,ab),
therefore d|a+b, d|a-b, d|ab
and also d|(a+b+a-b)=2a and d|(a+b-[a-b])=2b
So d|gcd(2a,2b)
but since gcd(a,b)=1 --> 2*gcd(a,b)=2 --> gcd(2a,2b)=2
so from this d|2 and so d=1 or d=2
from here it is the "ab" that is bugging me and will...