Turing recognizable language between languages that aren't recognizable?
The answer is no, of course.

Consider, for instance, the following languages:


$L_1 = \{ \langle M \rangle \# \langle M \rangle \mid \, \text{$M$ is a TM which does not halt on input $\langle M \rangle$} \}$
$L = \{ \langle M \rangle \# \langle M \rangle \mid \, \text{$M$ is a TM} \}$
$L_2 = L \cup \{ \langle M \rangle \# w \mid \, \text{$M$ is a TM which does not halt on input $w$} \}$


Then it is easy to see neither $L_1$ nor $L_2$ are recursively enumerable (i.e., Turing-recognizable), despite $L$ being even context-sensitive. (In fact, you could even make $L$ context-free by replacing the second $\langle M \rangle$ with its reversed copy.)



The upshot is that set inclusion (on its own) tells us very little about the sets involved being recursively enumerable or not, unless the symmetric difference between them is finite. In the case of $L_1 \subseteq L \subseteq L_2$, we do know, for instance, that $|L \setminus L_1|$ or $|L_2 \setminus L|$ being finite implies that $L_2$ is also not recursively enumerable.