I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see , and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);

as , it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

--Kevin C.