REC and RE under intersection
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.