$$L = \{ \langle M \rangle \mid \text{\(M\) is a Turing Machine and \(|L(M)| = 1\)} \}$$I have to prove that this is not R.E. and not co-R.E.I know how to approach these kind of problems. For$\check L = \{ \langle M \rangle \mid \text{\(M\) is a Turing Machine and \(|L(M)| \le 1\)} \}$,Here a simple reduction from complement of HP would do the trick. (Make a TM $M'$ which on input $w$ runs $M$ on $x$ and accepts if $M$ accepts $x$. If $\langle M,x \rangle$ belongs to complement of HP then $x$ would not be accepted and hence $M'$ would reject everything thereby satisfying the $\le 1$ property. Also in case $\langle M,x \rangle$ doesn't belong to complement of HP, $M'$ would accept everything and hence would be $\gt 1$. Thus we know that the complement of HP is reducible to $\check L$. Hence $\check L$ is not RE.)But in the main problem I don't have an inequality and I am stuck with how to approach such problems. I would face the same problem while proving $L'$ (complement of $L$) as not RE for proving $L$ as not co RE.
Proving a language is neither Recursively Enumerable nor co-Recursively Enumerable
turing machines;proof techniques;undecidability
One way to solve this problem is by showing there do not exist Turing machines that accept  $L_3$ or $\overline L$. But probably that is simply restating the question. $L_u$, the universal language, is non-recursive but RE language. If we reduce $L_u \leq_M L$ then $L$ cannot be co-RE. If we reduce $L_u \leq_M \overline{L}$ then $L$ cannot be RE.Let us  first do the reduction $L_u \leq_M L$.We shall reduce $\langle M,w\rangle$ to a Turing machine $M'$ as follows:$M'$ rejects all inputs other than $w$. If the input is $w$ then it runs as Turing Machine $M$ on $w$. This gives us $\langle M,w\rangle \in L_u$ iff $M' = f(\langle M,w\rangle) \in L$.Hence $L_u \leq_M L$. Therefore $L$ cannot be co-RE. Otherwise $L_u$ will become recursive.Next we do the reduction $L_u \leq_M \overline L$.We shall reduce $\langle M,w\rangle$ to a Turing machine $M''$ as follows: Let $w'$ be a string other than $w$.$M''$ rejects all inputs other than $w$ and $w'$. If the input in $w'$ then $M''$ accepts the input. If the input is $w$ then it runs as Turing Machine $M$ on $w$. This gives us $\langle M,w\rangle \in L_u$ iff $M''= f(\langle M,w\rangle) \not\in L$, that is, $\langle M,w\rangle \in L_u$ iff $M'' \in \overline L$.Hence $L_u \leq_M \overline{L}$. Therefore $L$ cannot be RE. Otherwise $L_u$ will become recursive.If we combine the results, $L$ is neither RE nor co-RE.