G1,G2 groups G2 not abelian then Hom(G1,G2)need not be even semigroup under composition fg(x)=f(x)g(x)true/fal

Answer 1

you said composition but you wrote multimplication( of functions).
You meant "operation" maybe
The multiplication in G2 is associative so the operation is associative:
f(gh) = (fg)h because
f(x)(g(x)h(x)) =(f(x)g(x))h(x) for any x in G1.
So Hom(G1,G2) is semigroup, it can be empty.