Outer measure of a union of 2 subsets of disjoint measurable sets of real numbers.

Put $F=E\cap (A\cup B)$ then, since $A$ is $m^*$-measurable, we have that $$ m^*(F)= m^*(F\cap A) + m^*(F\cap A^c) = m^*(E\cap A)+m^*(E\cap B) $$ since $A\cap B =\emptyset$

We start with a little Lemma:

Lemma. Let $E\subseteq \Bbb R$. If $H\supseteq E$ is a $G_\delta$ set (countable intersection of open sets) such that $$m(H)=m^\ast(E),$$ then for every $C\subseteq\Bbb R$ $$m^\ast(H\cap C)=m^\ast(E\cap C).$$

Proof. Let $C\subseteq\Bbb R$. In the following the superscript $^c$ means complement. $$\begin{align*} m^\ast(H\cap C) &\leq m^\ast(H\cap C\cap E\cap C)+m^\ast((H\cap C)\setminus (E\cap C))\\ &= m^\ast(E\cap C) + m^\ast((H\cap C)\cap (E\cap C)^c)\\ &= m^\ast(E\cap C) + m^\ast(C\cap (H\setminus E))\\ &\leq m^\ast(E\cap C) + m^\ast(H\setminus E)\\ &= m^\ast(E\cap C) \end{align*}$$ The inequality $m^\ast(H\cap C)\geq m^\ast(E\cap C)$ comes free by the monotony of the outer measure since $H\supseteq E$.

Proof of $m^\ast(E\cap (A\cup B))\geq m^\ast(E\cap A)+m^\ast(E\cap B)$.

Pick $H\supseteq E$ a $G_\delta$ set so that $m(H)=m^\ast(E)$. Then $$\begin{align*} m^\ast(E\cap (A\cup B)) &= m^\ast(H\cap (A\cup B)) &&\text{by the Lemma}\\ &= m(H\cap A) + m(H\cap B) &&\text{($^\ast$)}\\ &\geq m^\ast(E\cap A) + m^\ast(E\cap B) &&\text{by the monotony of the outer measure.} \end{align*}$$ ($^\ast$) because here we are dealing with measurable sets (of finite measure).

Observation. Notice that such a $G_\delta$ set $H$ always exist even if $E$ is unbounded.