REC and RE under intersection Would the intersection of a recursive language and a recursively enumarable language be recursive or recurisvely enumbarable or neither? 

Assume $L_{3}$ is the intersection of some language $L_{1}$ $\epsilon$ RE and some other language $L_{2}$ $\epsilon$ REC .

Since $L_{2}$ $\epsilon$ REC, there must be a Turing-Machine $M_{2}$ that holds on every input, whethere or not the input x $\epsilon$ $L_{2}$.
If we now have a k-NTM $M_{3}$ that on one band simulates 
$M_{2}$ and upon acceptance of the input switches to another band where it then simulates $M_{1}$ (for $L_{1}$).

Since $M_{1}$ is only ever simulated if $M_{2}$ accepts the input, $M_{1}$ only gets the chance to hold for every word x $\epsilon$ $L_{1}$ $\bigcap $ $L_{2}$.

That means we can already assume that we wont get stuck in a loop for words x $\notin $ $L_{2}$.

However that is not the case for words that are in the intersection or only in $L_{2}$ and not in $L_{1}$, as the simulation of $M_{1}$ wil also be recursively enumarable and may run forever, without us ever knowing whether the input is accepted or not.

If my thoughts so far were correct, that means that  $L_{3}$ $\epsilon$ RE.

What I am uncertain about is if there is any sort of algorithm or technique with which we could make the simulation of $M_{1}$ decidable, or if maybe it is possible to make an entirely new TM, independent from $M_{1}$ and $M_{2}$ and make sure it is decidable?
You have been going well. Yes, $L_3=L_1\cap L_2\in\text{ RE} $.

Could it be that $L_3$ is recursive?

The answer is not necessarily. There are several cases here.


If $L_1$ is a recursive language as well, then $L_3$ is also recursive.
If $L_1$ is not a recursive language, then $L_3$ may or may not be recursive.


For example, let $L_2$ be the languages of all words. Then $L_3=L_2$ is not recursive.
For example, let $L_2$ be the language that does not contain any word. Then $L_3$ is also the language that does not contain any word, which is recursive.





Exercise 1. Show that if both $L_1$ and $L_2$ are recursive, then $L_1\cap L_2$ is recursive. 

Exercise 2. Construct two languages $L_1$ and $L_2$ that satisfy all following conditions.


$L_1$ is recursively enumerable.  
$L_2$ is recursive and $L_2$ is not the language of all words.  
$L_1\cap L_2\subsetneq L_1$ is recursively enumerable but not recursive.