Eigenvalues and eigenspaces of almost complex structures under each other [closed]

The answer to all your questions is yes and has nothing to do with complexification. Let $V$ be a real vector space and let $J,H$ be two commuting linear complex structures on $V$ ($J^2 = H^2 = -\operatorname{id}_W$ and $JH = HJ$).

You can consider $V$ as a complex vector space with respect to $H$. Then, since $J$ commutes with $H$, the map $J$ is actually $\mathbb{C}$-linear as a map $J^H \colon (V,H) \rightarrow (V,H)$. As a real map, $J \colon V \rightarrow V$ doesn't have any eigenvalues since if $Jv = \lambda v$ then $J^2v = \lambda^2 v = -v$ which implies that $\lambda^2 = -1$. Considering $J^H$ as a complex linear map, the above calculation shows that the only possible eigenvalues of $J^H$ are $\pm i$. We also have a direct sum decomposition

$$ V = \{ v \in V \, | \, J^H v = iv \ \iff Jv = Hv \} \oplus \{ v \in V \, | \, J^Hv = -iv \iff Jv = -Hv\} $$

where the first factor is the "eigenspace" of $J^H$ corresponding to the eigenvalue $i$ and the second is the "eigenspace" of $J^H$ corresponding to the eigenvalue $-i$. The only caveat is that one of the factors might be trivial so $J^H$ won't necessarily have both $\pm i$ as eigenvalues.

Similarly, you can consider $V$ as a complex vector space with respect to $J$ and then $H^J \colon (V,J) \rightarrow (V,J)$ is $\mathbb{C}$-linear with the only possible eigenvalues being $\pm i$ and you get a direct sum decomposition

$$ V = \{ v \in V \, | \, H^Jv = iv \iff Hv = Jv \} \oplus \{ v \in V \, | \, H^Jv = -iv \iff Hv = -Jv \} $$ where the first factor is the "eigenspace" of $H^J$ corresponding to the eigenvalue $i$ and the second is the "eigenspace" of $H^J$ corresponding to the eigenvalue $-i$. This shows that $J^H$ and $H^J$ have the same eigenvalues and the same eigenspaces.

Finally, the map $J^J \colon (V,J) \rightarrow (V,J)$ is also complex linear and is just given by multiplication by $i$ so it has only $i$ as an eigenvalue (at least as long as $V \neq \{ 0 \}$).