If $E$ is a measurable set, how to prove that there are Borel sets $A$ and $B$ such that $A\subset E$, $E\subset B$ and $m(A)=m(E)=m(B)$? If $E$ is a measurable set, then  how to prove that there are Borel sets $A$ and $B$ such that $A$ is a subset of $E$, $E$ is a subset of $B$ and $m(A)=m(E)=m(B)$?

I assume by $m$ you mean Lebesgue measure on $\mathbb R^n$. Use that this measure is regular. This gives us that, if $m(E)<\infty$, then for any $n$ there are a compact set $K_n$ and and open set $U_n$ with $K_n\subset E\subset U_n$, and $m(E)-1/n<m(K_n)$ and $m(U_n)<m(E)+1/n$.
This implies that $A=\bigcup_n K_n$ and $B=\bigcap U_n$ have the same measure as $E$, and they are clearly a Borel subset and a Borel superset of $E$, respectively.
If $E$ has infinite measure, it is even easier: Take as $B$ the set $\mathbb R^n$. As before, regularity gives us for each $n$ a compact set $K_n$ with $K_n\subset E$ and $m(K_n)\ge n$. Then we can again take as $A$ the set $\bigcup_n K_n$. 
