1. Using dovetailing: (intuitive and best)
Property : Acceptance of a finite language $\begin{align} L(M) = \phi \cup \left \{ 00 \right \}\cup \left \{ 11 \right \}\cup \left \{ 00,11 \right \} \\ \end{align}$.
For checking any $M$ belonging to the property, we have to use an UTM and check $\langle M,w \rangle$ pairs , feeding $w$ in a dovetailing (zigzag) manner.
We have to confirm the acceptance of some strings (as shown above) and non-acceptance of all other strings.
"non-acceptance of all other strings", this requirement is hard and TM may go infinite loop.
So we can not even check $M$ for although $M$ may belong to property.
$\Rightarrow$ $L1$ is non-computable or NOT-RE.
2. Using Rice Theorem.
$$\begin{align} T_{1_{yes}} &= \phi = \left \{ \right \} \\ T_{2_{yes}} &= \left \{ 00 \right \}
\\ T_{3_{yes}} &= \left \{ 11 \right \}
\\ T_{4_{yes}}
&= \left \{ 11,00 \right \} \\ \end{align}$$
Whenever $\phi \in \text{Property SET}$ then corresponding property language is NOT-RE.
Or, Some supersets of any $T_{yes}$ does not satisfy the given property. Hence NOT RE