Prove that union, intersection and difference of measurable sets is measurable using specific definition of measurability

I found a lot of questions with the similar title on the forum, but not all of them are properly answered and those, which are, differ significantly from mine.

I work in the closed interval $[0,1]$. I use the following definition of Lebesgue measurability:

$\mathbf{Def. 1}$ The set E is called measurable iff $$\lambda^{*}(E)+\lambda^{*}([0,1]\backslash E)=1,$$ where $\lambda^{*}$ is the outer lebesgue measure, i.e. $\lambda^{*}(E)=\inf_{U\supset E}\lambda(U)$, where $U$ is an open set.

I can prove that this definition is equivalent to the following:

$\mathbf{Def. 2}$ The set E is called measurable iff $\forall \epsilon >0 \ \exists U_{\epsilon} ,F_{\epsilon}$ - open and closed sets respectively, such that $F_{\epsilon} \subset E \subset U_{\epsilon}\ \lambda^{*}(U_{\epsilon})-\lambda^{*}(F_{\epsilon})<\epsilon$.

$\sigma$-additivity for $\lambda$ is also proven (for a measurable set $E$: $\lambda(E)=\lambda^{*}(E))$ :

If $\{E_n\}$ - a collection of measurable sets and $E_k\cap E_l = \varnothing$ then $\lambda \left (\bigcup_{n=1}^{\infty}E_n \right)=\sum_{n=1}^\infty{\lambda(E_n)}$.

And here is my question:

a) If $A$ and $B$ are measurable sets (they may intersect) prove that $A\cup B, A\cap B, A\backslash B$ are measurable using $\mathbf{ Def. 1}$ or $\mathbf{ Def. 2}$.

b) Prove that countable intersection and countable union of the measurable sets is measurable.

Usually, a set $E$ is called (Lebesgue) measurable, if for all sets $A$ we have $$\lambda^*(E\cap A)+\lambda^*(E^\complement\cap A)=\lambda^*(A).$$ We can show, that Def. 1 is equivalent to this standard definition of Lebesgue measurability. Then a) and b) follow, (you can read their proofs in a book on measure theory, they are a bit long, though).
Proof: If $E$ is Lebesgue measurable (by the standard definition), then it satisfies the equality in Def. 1 (take $A=[0,1]$).
Let $E$ be not Lebesgue measurable. Then there is a set $A$ and an $\varepsilon>0$ such that $$\lambda^*(E\cap A)+\lambda^*(E^\complement\cap A)>\lambda^*(A)+4\varepsilon.$$ By the definition of $\lambda^*$ there is a countable collection of open intervals $(I_n)_n$ where $I_n=(a_n,b_n)$ which together cover $A$ and where $$\lambda^*(J)\leq\sum_{n=1}^\infty(b_n-a_n)<\lambda^*(A)+2\varepsilon,$$ for $J=\bigcup_nI_n$. Since $A\subseteq J$ also $\lambda^*(E\cap J)+\lambda^*(E^\complement\cap J)>\lambda^*(A)+2\varepsilon$. Since the tail of the above sum goes to $0$, there is an index $N$ such that $$\sum_{n=N}^\infty(b_n-a_n)<\varepsilon.$$ Let $J_N=\bigcup_{n=1}^NI_n$. Then $$\lambda^*(E\cap J)\leq\lambda^*(E\cap J_N)+\lambda[E\cap(J\setminus J_N)]<\lambda^*(E\cap J_N)+\varepsilon$$ and the same for $E^\complement$ instead of $E$. The next thing works, because $J_N$ and $J_N^\complement$ are finite unions of intervals and disjoint. \begin{align} \lambda^*(E)+\lambda^*(E^\complement)&=\lambda^*[(E\cap J_N)\cup(E\cap J_N^\complement)]+\lambda^*[(E^\complement\cap J_N)\cup(E^\complement\cap J_N^\complement)]\\ &=\lambda^*(E\cap J_N)+\lambda^*(E\cap J_N^\complement)+\lambda^*(E^\complement\cap J_N)+\lambda^*(E^\complement\cap J_N^\complement)\\ &>[\lambda^*(E\cap J)+\lambda^*(E^\complement\cap J)-2\varepsilon]+\lambda(J_N^\complement)\\ &>\lambda^*(A)+4\varepsilon-2\varepsilon+[1-\lambda(J)]\\ &>1,\end{align} which means, that $E$ is neither Lebesgue measurable according to Def. 1.