Decidability of languages containing bitstrings and their corresponding ones' complements

Consider strings $s \in \{0,1\}^*$. Define $c_1(s)$ to be the ones' complement of $s$; i.e., the string obtained from $s$ by inverting all of its bits. So, for example, $c_1(000111) = 111000$. Call a language $L \subset \{0,1\}^*$ ones' complement closed, or OCC, if $s' \in L$ $\iff$ $c_1(s') \in L$, provided $s'$ is not the empty string. Given these assumptions, I have a few questions.

Does there exist an OCC language that is both Turing-recognizable and not decidable?

Does there exist an OCC language $L$ with the property that both $L$ is not decidable and $\overline{L}$ is Turing-recognizable?

Does there exist an OCC language $L$ that is such that neither it nor $\overline{L}$ is Turing-recognizable?

I am having some trouble answering these questions because of how limited the alphabet $\{0,1\}$ is. I was thinking that perhaps this alphabet alone could be used to encode Turing machines, and from there we could talk about the well-known decidable/undecidable and recognizable/not-recognizable languages involving Turing machine encodings, but I'm not sure this is possible.

Given a language $L$, you can create a language $L'$ which is equivalent in power and OCC: $$L' = \{ 0 x : x \in L \} \cup \{ 1 c_1(x) : x \in L \}.$$ The two languages are recursively equivalent. This means that there is a computable reduction from $L$ to $L'$ and another one from $L'$ to $L$.