I am interested in the classic problem REGULAR LANGUAGE INCLUSION. Given a regular expression $E$, we denote by $L(E)$ the regular language associated to it. (Regular expressions are on a fixed alphabet $\Sigma$, with the operations union, Kleene-star and concatenation.)
Input: Two regular expressions $E_1$ and $E_2$
Question: Is it true that $L(E_1)\subseteq L(E_2)$?
REGULAR LANGUAGE INCLUSION is known to be PSPACE-complete .
The classic way to solve it (in PSPACE) is to construct the NFAs $A_1$ and $A_2$ associated to $E_1$ and $E_2$, to build a DFA $D_2$ from $A_2$, complement it into a DFA $D_2^C$, and finally, build the intersection automaton $A_P$ from $A_1$ and $D_2^C$ corresponding to the intersection of $L(E_1)$ and $L(E_2)^C$. Now $L(E_1)\subseteq L(E_2)$ if and only if there no an accepting path in $A_P$.
If I am not mistaken, the whole process can be done in polynomial time when $E_2$ is a fixed language, since the only exponential blow-up comes from transforming $A_2$ into $D_2$. Even better, the problem is FPT when parameterized by $|E_2|$, the length of $E_2$.
This motivates my question:
Question: When $E_1$ is a fixed expression, what is the complexity of REGULAR LANGUAGE INCLUSION? Does it remain PSPACE-complete?
 L. J. Stockmeyer and A. R. Meyer. Word problems requiring exponential time: preliminary report. Proceedings of the fifth annual ACM Symposium on Theory of Computing, STOC '73, pp. 1-9.
Remark: Being a non-expert in the field, I find  (and related papers of that time) quite unreadable, and couldn't find another proof of the PSPACE-completeness - any pointer to a modern proof, such as in a book, is very welcome! Also, the authors seem to allow squaring in their regular expressions, which is nowadays rather non-standard, I believe.)