Union of two non-context-free languages Let L1 = L2 union L3
find values such that L1 is context free and L2 and L3 are not.

So far I have:
L1 = $a^nb^n$
L2 = $a^*b^*$
L3 = $a^+b^+$

Is this acceptable?? Since L2 covers everything including epsilon and L3 is the same but does not include epsilon?

I know L2 is regular so I guess that is also not a CFL.
Another problem is that the a's and b's aren't linked in L2 and L3, so either one can always have more a's than b's and vice versa.
Hint. For any language $L$, $$A^* = L \cup (A^* - L).$$
Now, choose an appropriate language $L$ to solve your problem.