Reduction between $\Sigma^*$ and $\emptyset$
Let $f : \Sigma^* \to \Sigma^*$ any total function. Then, the statement $x \in L_2$ is always false since $L_2 = \emptyset$, and conversely $f(x) \in L_1$ is always true since $L_1 = \Sigma^*$. So, clearly, the equivalence

$\qquad\displaystyle x \in L_2 \iff f(x) \in L_1$

is false for $f$ and all $x \in \Sigma^*$ (it would have been sufficient to find one example-$x$); in fact,

$\qquad\displaystyle x \in L_2 \iff f(x) \notin L_1$

holds for all $x$. So $f$ is not a reduction function.

Since we picked $f$ arbitrarily, there can be no reduction from $L_2$ to $L_1$. A similar argument works for the other direction.