How to prove that the reduction relation is not symmetric
In order to show that $L_1 \leq L_2$, we need to come up with a computable function $f$ such that $x \in L_1$ iff $f(x) \in L_2$. Since $L_1 = \Sigma^*$, every $x$ satisfies $x \in L$, and so we need to find a computable function $f$ such that $f(x) \in L_2$ for all $x$. The easiest way to satisfy this is to choose $f$ to be a constant function.
In order to show that $L_2 \not\leq L_1$, we need to show that $f$ is a computable function that it is not the case that $x \in L_2$ iff $f(x) \in L_1$. That is, we need to show that for every computable $f$ there exists $x$ such that it is not the case that $x \in L_2$ iff $f(x) \in L_1$. Since $L_1 = \Sigma^*$, we always have $f(x) \in L_1$, so the only way in which &quot;$x \in L_2$ iff $f(x) \in L_1$&quot; can fail is if $x \notin L_2$. That is, we need to show that for every computable $f$ there exists $x$ such that $x \notin L_2$. The goal doesn't involve $f$, so it suffices to show that there exists $x$ such that $x \notin L_2$.
You take it from here.