Is at least one of L and L complement Turing recognizable
No, try to think of a counting argument (i.e, the size of each class) in the following way:
There are $2^{\aleph_0}$ possible words in the language, hence $2^{2^{\aleph_0}}$ total languages. However, the number of turing machines is bounded by the total possible number of words (since you can encode a TM as a string), hence there are at most $2^{\aleph_0}$ TMs, and thus we get that $|RE| \le 2^{\aleph_0}$. Follow the definition of $co-RE$, to see that the same argument gives $|co-RE| \le 2^{\aleph_0}$ as well. Thus, $|RE\cup co-RE| \le 2^{\aleph_0}$ as well.
Thus $|RE\cup co-RE|\le 2^{\aleph_0}&lt;2^{2^{\aleph_0}}=|P(\{0,1\}^*)|$, and thus there must be some $L\in P(\{0,1\}^*)\setminus (RE\cup co-RE)$ as you asked.
In fact, this shows there are waaaaay more languages that aren't RE or co-RE than languages that are.