Both $L$ and $Lʼ$ are undecidable and not even semi-decidable (not recursively-enumerable). Because halting problem can be solved with both $L$ and $Lʼ$.

Halting problem can be stated as follows: A machine $M$ and a word $w$ are given. You have to tell, if $M$ halts on $w$.

So, to solve halting problem $\langle M,w\rangle$ using $L$, just give $\langle M,w,0\rangle$ and $\langle M,w,1\rangle$ to two instances of $T$ which is the assumed Turing machine for $L$. If $T$ accepts the triplet $\langle M,w,0\rangle$, it means $M$ halts on $w$ => we have solved halting problem. If $T$ accepts the triplet $\langle M,w,1\rangle$, it means $M$ doesn't halt on $w$ => we have solved halting problem. We know that either $\langle M,w,0\rangle$ or $\langle M,w,1\rangle$ is in $L$. So, if $L$ is recursively enumerable, $T$ is bound to stop on at least one of these inputs ($TM$ for a recursively enumerable language stops and accepts, when provided with a word in its language).

Hence, if $L$ is recursively enumerable we can solve halting problem $\Rightarrow L$ is not recursively enumerable.

Similarly, we can also show that halting problem can be solved with $L'$. (shown at end)

Hence, neither $L$ nor $L'$ is recursively enumerable.

To solve halting problem $\langle M,w\rangle$ using $L'$, just give $\langle M,w,0\rangle$ and $\langle M,w,1\rangle$ to two instances of $T'$ which is the assumed Turing machine for $L'$. If $T'$ accepts the triplet $\langle M,w,0\rangle$, it means $M$ does not halt on $w$ => we have solved halting problem. If $T$ accepts the triplet $\langle M,w,1\rangle$, it means $M$ halt on $w$ => we have solved halting problem. We know that either $\langle M,w,0\rangle$ or $\langle M,w,1\rangle$ is in $L'$. So, if $L'$ is recursively enumerable, $T'$ is bound to stop on at least one of these inputs ($TM$ for a recursively enumerable language stops and accepts, when provided with a word in its language).

Hence, if $L'$ is recursively enumerable we can solve halting problem $\Rightarrow L'$ is not recursively enumerable.

PS: If the bit part of the triplet is absent then $L_0$ is halting problem and $L_1$ is its complement and $L_0 \cup L_1 = \Sigma^*$, which is regular. Lets see how it happens.

Let the alphabet set be $\{0,1\}$. Now for any string like $0010101$ there are only two options- belong to $L$ or belong to $L'$ as this is what complement says. Now, lets take the case for $L_0$ and a string $001\dots10-01-1$, ("-" shown for notation purpose only) where the first component describes a TM $M$ followed by input "$w = 01$" and last bit "1". Now suppose $M$ halts on "$01$". Still the given input is not in $L_0$ as the last bit is "$1$" and not "$0$" as required by $L_0$. So, this input must be in $L_0'$. But since $M$ halts on $w$, this input is not in $L_1$ either. Similarly, we can get an infinite set of strings which does not belong to both $L_0$ and $L_1$ and this makes their union not $\Sigma^*$ but an irregular (not r.e. as proved earlier) set. If the last bit is removed from the definition of $L_0$ and $L_1$, then any string should be present in either $L_0$ or $L_1$ and their union would be $\Sigma^*$.