1.26

It's fairly easy to see that since $B = A \cup (B \backslash A)$, that since both $A$ and $B$ are measurable, $m(B) = m(A) + m(B \backslash A)$. Thus, $m(B \backslash A) = 0$. Similarly, we observe $m^{*}(E \backslash A) = 0$, and thus $E \backslash A$ is measurable, with measure zero.

Certainly, $E = A \cup (E \backslash A)$. Thus, since $A$ is measurable, and the union of any two measurable sets is measurable, we have $E$ must be measurable.