Let $\mu$ be a finite measure on $(X, M)$, and let $\mu^*$ be the outermeasure induced by $\mu$. Suppose that $E \subset X$ satisfies $\mu^*(E) = \mu^*(X)$ (but not that $E \in M$).
a.) If $A,B \in M$ and $ A \cap E = B \cap E$, then $\mu(A) = \mu(B)$. b.) Let $M_E = \{A \cap E : A \in M \}$ , and define the function $v$ on $M_E$ defined by $v(A \cap E)= \mu(A)$ (which makes sense by (a)). Then $M_E$ is a $\sigma$-algebra and $v$ is a measure on $M_E$.
I am not sure how to start either of these problems. $E \notin M$ is throwing me off.
The key point in a) is that $\mu^*(E)=\mu(X)$. So $\mu^*(E^c)=0$. Then $$ \mu^*(A)\leq\mu^*(A\cap E)+\mu^*(A\cap E^c) \leq\mu^*(A\cap E)+\mu^*(E^c)=\mu^*(A\cap E)=\mu^*(B\cap E)\leq\mu(B). $$ As the roles of $A$ and $B$ can be reversed, you get $\mu(A)=\mu(B)$.
For b): that $M_E$ is a $\sigma$-\algebra follows from the fact that $M$ is a $\sigma$-algebra. You have $E=X\cap E\in M_E$. And $(A\cap E)\cap(B\cap E)=(A\cap B)\cap E$, so $M_E$ contains finite intersections. Similarly, $$\bigcup_n(A_n\cap E)=\left(\bigcup_n A_n\right)\cap E.$$ To check that $v$ is a measure, if $\{A_n\cap E\}$ are disjoint, then $$ v\left(\bigcup_nA_n\cap E\right)=v\left(E\cap\bigcup_n A_n\right)=\mu\left(\bigcup_nA_n\right)=\sum_n\mu(A_n)=\sum_nv(A_n\cap E). $$