Two languages $L,\bar{L}$
1 : Both are REC : Yes, It is possible. Since Set of REC languages is closed under complementation operation.
2 : Both are RE : Yes, It is also possible. Every REC language is RE and Since Set of REC languages is closed under
complementation operation.
3 : One is REC & other is RE : Yes, It is possible. If One($L$) is REC, then other($\bar{L}$) will be REC and so RE (as Set of REC languages is closed under complementation operation.)
4 : Both are RE BUT NOT REC : No, It is Not possible. There is Theorem which states that " If a language and its complement are both recursively enumerable, then both are recursive." (Proof given at the end of the answer) So, Both are RE but Not REC is Never possible.
5 : One is REC & other is NOT RE : No, Not possible. If One($L$) is REC, then other($\bar{L}$) will also be REC and so RE (as Set of REC languages is closed under complementation operation.)
6 : One is RE & other is RE BUT NOT REC : No, Not possible. By the theorem stated in the 4th point, We can easily derive that the Complement of "RE but Not REC" is Always "Not RE".
7 : One is NOT RE & other is RE : Yes, It is Possible. One($L$) is Not RE, So, $\bar{L}$ will be either "Not RE" Or "RE but Not REC".
Theorem: If a language and its complement are both recursively enumerable, then both are recursive.
Proof:
Suppose a language $L$ is recursively enumerable. That means there exists a Turing machine $T1$ that, given any string of the language, halts and accepts that string.
Now let's also suppose that the complement of $L$, $\bar{L}$ $=$ {$w: w \notin L$}, is recursively enumerable. That means there is some other Turing machine $T2$ that, given any string of $\bar{L}$, halts and accepts that string.
Clearly, any string (over the appropriate alphabet ) belongs to either $L$ or $\bar{L}$. Hence, any string will cause either $T1$ or $T2$ (or both) to halt. We construct a new Turing machine that emulates both $T1$ and $T2$, alternating moves between them. When either one stops, we can tell (by whether it accepted or rejected the string) to which language the string belongs.
Thus, we have constructed a Turing machine that, for each input, halts with an answer whether or not the string belongs to $L$. Therefore $L$ and $\bar{L}$ are recursive languages.
