Can the following conditions hold all together?
Conditions $2,3,4$ imply $L_1 \in RE$, contradicting $1$.

Here's how to semi-decide $L_1$. Take a candidate $x$.
If $x \in L_1 \cap L_2$ (decidable by $2$), accept.
If $x \notin L_1 \cup L_2$ (decidable by $3$), diverge (we know $x \notin L_1$ for sure).

We now know that
$$x \in (L_1\setminus L_2) \cup (L_2\setminus L_1) \qquad (*)$$

Exploiting $4$, we run semi-deciders for $\overline{L_1}, \overline{L_2}$ on $x$ in parallel. By $(*)$, (exactly) one of these must eventually halt.

If the former halts, we infer $x \notin L_1$, so we diverge. 
If the second one halts, we infer $x \notin L_2$, and by $(*)$ we obtain $x\in L_1$, so we accept.