Prove if $E_1$ and $E_2$ are measurable then $m(E_1\cup E_2)+m(E_2\cap E_2)=m(E_1)+m(E_2)$ by additivity
$m(E_1\cup  E_2)=m(E_1)+m(E_2)$ (because $E_1,E_2$ are measurable)
but i don't know what to do with $E_1\cap E_2$.
I tried to use demorgan's identity to solve this part but this is not clear to me.  
 A: Firstly, $m(E_1 \cup E_2) = m(E_1) + m(E_2)$ is not true in general. Its only true if $E_1 \cap E_2 = \emptyset$.
So, with that cleared up, note that there are sets $A, B$ and $C$ such that $A\cap B = A\cap C = B\cap C = \emptyset$ and $E_1 = A \cup B,\ E_2 = B\cup C$ and $E_1\cap E_2 = B$. 
Therefore $E_1\cup E_2 = A\cup B\cup C$ and using the fact that we have additivity for disjoint sets, we can conclude $$\begin{align*}m(E_1 \cup E_2) + m(E_1\cap E_2) &= m(A\cup B\cup C) + m(B)\\
&= m(A) + m(B) + m(C) + m(B)\\
&= m(A\cup B) + m(B\cup C)\\
&= m(E_1) +m(E_2).\end{align*}$$
So, big important thing to remember, what you have in general is$$ m\left(\bigcup_i A_i\right) \leq \sum_i m(A_i)$$with equality only holding in the case of disjoint sets.
A: Hint: $E_1\cup E_2 = (E_1-(E_1\cap E_2))\cup (E_2-(E_1\cap E_2)) \cup (E_1 \cap E_2)$, when the unions in the RHS are disjoint.
Did that help? Can you use the claim you already stated in the question?
A: We have $E_1\setminus (E_1 \cap E_2) $  and  $E_2$ are disjoint, then  $$ m(E_1 \cup E_2)= m( E_1\setminus (E_1 \cap E_2))  +  m(E_2)$$
adding $m(E_1 \cap E_2)$ on both sides we get 
$$m(E_1 \cup E_2)+ m(E_1 \cap E_2) = m( E_1\setminus (E_1 \cap E_2))  +  m(E_2) +m(E_1 \cap E_2) $$
But $E_1 \cap E_2 $ and  $E_1\setminus (E_1 \cap E_2) $  are also disjoint so
 $m(E_1 \cap E_2) + m(E_1\setminus (E_1 \cap E_2)) = m((E_1 \cap E_2) \cup (E_1\setminus (E_1 \cap E_2))) =m(E_1)$ , which implies
$$m(E_1 \cup E_2)+ m(E_1 \cap E_2) = m( E_1)  +  m(E_2) $$
