Self doubt decidability

Question

$\text{L}_1 = \{\text{<M>}|\text{M is a TM and L(M) is infinite} \}$

Not RE

$\text{L}_2 = \{ \text{<M>}  | \text{M is a TM and L(M) is countable} \}$

and 

$\text{L}_3 = \{ \text{<M>}  | \text{M is a TM and L(M) is uncountable}\}$

both are Recursive

how? 

Comments

thanks but try to apply same answer to countable infinite language why it becomes recursive?

---

Mk Utkarsh please explain a bit I am not getting your point :)

---

For L2 and L3 we cannot come with cases like "yes" case or "no" case . Even if you have to prove a set of languages is not countable, we make use of diagonalization. So we cannot come up with any example of "no" case making it recursive.