@dtldarek I have a question about the following:

If $A,B$ are recursively enumerable, then deduce if the sets $A \setminus B$, $A \triangle B=(A \cup B) \setminus{(A \cap B)}, AB, A \cap B$ are also recursive.

@dtldarek The halting problem is recursive enumerable, but not decidable.

No, its complement isn't RE, otherwise it would be recursive.

@dtldarek You mean that we can reduct the halting problem $H=\{ (n,x)| T_n(x) \downarrow \}$ to $A \setminus{B}$ ?
So by taking as $A$ to be all the languages, we have that $A \setminus{B}=B^c$.

$B^c$ is the set $\{ x \in A: x \notin B\}$. How can we reduct $H$ to $B^c$ ?

Great. I got it... Thanks a lot!!! :) @dtldarek
Can we also use it for example in this case $A \triangle B=(A \cup B) \setminus (A \cap B)$, where $A,B$ recursively enumerable?

Then we have that $A \cup B= \Sigma^{\star}, A \cap B=H$, then $A \triangle B=\Sigma^{\star} \setminus{H}=H^c$ which is not recursively enumerable.

So $A \cap B$ is not recursive, right?

Oh, I am sorry. I wanted to write $(A \cup B) \setminus{A \cap B}=H^c$ is not RE, and so also not decidable.

But like that we also show that $A \cap B=H$ is not decidable since $H^c$ is not RE. Right?