If $A \cup B$ is measurable and $m(A \cup B) = m^*(A) + m^*(B) < \infty$ then A and B are measurables. That's it. I've only been able to find that, since $A \cup B$ is measurable: $$m^*(A) + m^*(B) = m(A \cup B) = m_*(A \cup B) \geqslant m_*(A) + m_*(B)$$
Maybe using too that if C is measurable and D is a subset of C then:
$$m(C)=m_*(D) + m^*(C \setminus D) $$
Maybe if i say this? $$m^*(A \cup B) = m^*(A)+m^*(B)-m^*(A \cap B) = m^*(A) + m^*(B) \Rightarrow$$ $$ \Rightarrow m^*(A \cap B)=0 \Rightarrow A\cap B = \varnothing$$
And now use this to say that: $$m(A \cup B) = m_*(A) + m^*((A\cup B)\setminus A) = m_*(A) + m^*(B) = m^*(A) + m^*(B) \hspace{2cm} \Rightarrow m_*(A)=m^*(A) < \infty$$
And, doing the same for B, I would have the result.
 A: Just need to prove that $A$ is measurable, and then using the same way, $B$ is Lebesgue measurable can be proved.
Firstly find a $G_{\sigma}$ set $H$ such that $B\subset H$ and $m(H)=m^{\ast}(B)$, and then let $E=(A\cup B)\cap H^c$. It is clear that $E$ is a measurable subset of $A$ (due to $E$ is also equal to $A-H$). Since
\begin{align*}
m^{\ast} (B)\leq m((A\cup B)\cap H)\leq m(H)=m^{\ast}(B)
\end{align*}
that is, $m^{\ast}(B)=m((A\cup B)\cap H)$. In addition, since 
\begin{align*}
m^{\ast}(A)+m^{\ast}(B)&=m(A\cup B)\\
&=m((A\cup B)\cap H^c)+m((A\cup B)\cap H)
\end{align*}
then 
\begin{align*}
m^{\ast}(A)=m((A\cap B)\cap H^c)
\end{align*}
After that, for any given $\epsilon>0$, there exists open set $G$ with $A\subset G$, such that $m(G)\leq m^{\ast}(A)+\epsilon$, and also exist closed set $F$ with $F\subset E$, such that $m(E-F)<\epsilon$. Then for $F\subset E\subset A\subset G$, 
\begin{align*}
m(G)< m^{\ast}(A)+\epsilon&=m(E)+\epsilon\\
&=m(E-F)+m(F)+\epsilon
\end{align*}
Thus, 
\begin{align*}
m(G-F)=m(G)-m(F)\leq m(E-F)+\epsilon <2\epsilon
\end{align*}
which implies that $A$ is Lebesgue measurable (Based on the equivalent condition of Lebesgue measurable). Similarly, $B$ is also Lebesgue measurable which can be proved by same way. 
A: Here is a solution without using Carathéodory's condition.
Find $G_\delta$ sets $H$ and $G$ such that $G \supset A$, $H \supset B$, $m(H) = m^*(B)$ and $m(G) = m^*(A)$. Set $U = A \cup B$.
Note that $m(G \cup H) \geq m(U)$ since $G \cup H \supset U$.
On the other hand, $m(G \cup H) \leq m(G) + m(H) = m^*(A) + m^*(B) = m(U)$.
So $m(G \cup H) = m(U) < \infty$.
Since $\infty > m(G \cup H) = m(G) + m(H) - m(G \cap H)$,
we have $m(G \cap H) = 0$.
By $A \subset G \cap U$, $m^*(A) \leq m(G \cap U) \leq m(G) = m^*(A)$.
So $m(G \cap U) = m^*(A)$.
Finally, $$\begin{align*}
m^*(G \setminus A) &\leq m^*(G \setminus A \cap B) + m^*(G \setminus A \setminus B)\\
&\leq m^*(G \cap H \setminus A) + m(G \setminus U) \\
&\leq m(G \cap H) + m(G) - m(G \cap U) = 0 . \end{align*}$$
So $A = G \setminus (G \setminus A)$ is Lebesgue measurable.
