Kleene closure, concatenation problem
For any language $L$, $$\emptyset⋅L=\emptyset$$.

Therefore, 
$$L_1\cdot L_2^* = \emptyset \cdot L_2^* = \emptyset$$

And thus $$L_1\cdot L_2^*\cup L_1^* = \emptyset \cup L_1^*= L_1^* = \{\epsilon\}$$.