What are the complexities of the following SAT subsets ? Assume $P \neq NP$

Let use the following notation ${}^ia$ for tetration (ie. ${}^ia = \underbrace{a^{a^{\cdot^{\cdot^{\cdot^{a}}}}}}_{i \mbox{ times}}$).

|x| is the size of the instance x.

Let L be a language, $L|_{f(i)\leq |x| &lt; g(i)} := \{ x \in L \mbox{ |  } \exists i \in \mathbb{N}\mbox{, } f(i) \leq |x| &lt; g(i) \}$

What is the complexity of the following languages :

$L_1 = SAT|_{{}^{2i}2 \leq |x| &lt; {}^{2i+1}2}$
$L_2 = SAT|_{{}^{2i+1}2 \leq |x| &lt; {}^{2i+2}2}$

As $L_1 \sqcup L_2 = SAT$, they can't be both in P under the assumption that $P \neq NP$.
As there both have exponential holes, I don't think SAT can be reduced to one.

Hence the intuition would be that they are both in NPI, but I can't find a proof or disproof.

Two others languages are
$L_3 = SAT|_{|x|={}^{2i+1}2}$
$L_4 = SAT|_{|x|={}^{2i}2}$

If one of both is in NPC, the other is in P because for each instance of one, it can't be transformed into an greater instance of the other because it is of exponential size, and smallers instances have a logarithmic size. Still by intuition, there is no reason why they would have a different complexity. What would their complexity be ?

Ladner's proof of NPI problems under $P \neq NP$ assumption use languages like $L_1$ or $L_2$, but $L_1$ and $L_2$ aren't built by diagonalization.
I think both are NPI under the stronger assumption (but obviously true) that NP is not in "infinitely often P" - i.e., every polynomial time algorithm A and every sufficiently large n, A fails to solve SAT on inputs of length n.

In this case, such languages are not in P, but they also cannot be NP complete, since otherwise a reduction from SAT to a language L with large holes will give an algorithm for SAT that succeeds on these holes.

Such an assumption is also necessary, since otherwise the languages can be in P, or NP-complete, depending on where the "easy input lengths" are located.