For an outer measure $m^*$, does $m^*(E\cup A)+m^*(E\cap A) = m^*(E)+m^*(A)$ always hold?
Solution 1:
$E$ being measurable gives, that for all $B$ $$ m^*(B) = m^*(B \cap E) + m^*(B \cap E^c)$$ applying this with $B = A \cup E$ gives $$ m^*(A \cup E) = m^*(E) + m^*(A \cap E^c) \iff m^*(A \cap E^c) = m^*(A \cup E) - m^*(E) \tag 1$$ and with $B = A$ $$ m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)\iff m^*(A \cap E^c) = m^*(A) - m^*(A \cap E) \tag 2$$ Now equate (1) and (2).