For any homomorphism f, we must have either f(1)= 0 or or f(1)= 1. Look at f(n)= nf(1) and nf(m/n)= f(n(m/n))= mf(1).

For all x in A and y in B, x< sup A and b< sup B so xy< sup A.sup B. Thus, sup A.sup B is an upper bound on A.B. Suppose there were an upper bound on A.B less than supA.supB. Show that there must be an upper bound on A less than sup A or an upperbound on B less than sup B.2. Let A, B be sets of positive reals numbers. Letīs define A.B={x.y; x belong to A and y belong to B}. Prove that if A and B are limited, then A.B is limited, where supreme (A.B)=suprmeA.supremeB and inf(A.B)=infA.infB

By the way, since I note you are from Costa Rica, some notes on English:

You "need help", not "needahelp", the word, in English, is "homomophism", the phrase is "either... or...", not "or ... or...", and sets are "bounded" not "limited".

Still, your English is far better than my Spanish!