Are there open problems for primes which are known for probable primes? Define "probable prime" (PP) to be natural $n>1$ satisfying $2^{n-1} \equiv 1 \pmod{n}$ or $n=2$.
Probable primes are the union of the primes and base two pseudoprimes.
This definition is much simpler than the definition for primes
and the primes are sufficiently large subset of probable primes.

Are there open problems for primes which are known for probable
  primes?

Positive answer doesn't necessarily mean the problem is
solved for the primes (e.g. infinitely many twin PP hypothetically might
mean finitely many twin primes and infinitely many twin base 2 pseudoprimes).
 A: The question of existence of large prime Fermat numbers $2^{2^n}+1$ is open, but the corresponding question for probable primes is straightforward to solve. $2^{2^n} \equiv -1 \mod 2^{2^n}+1$, so by squaring, we see that $2^{2^j}\equiv 1 \mod 2^{2^n}+1$ for all $j>n$.  In particular, $2^n > n$, so we get a positive answer for all non-negative integers $n$.
This gives another way to deterministically generate probable primes, and as opposed to the iterated Mersenne method, we get a sequence that grows only doubly exponentially instead of by a "tower of exponentials".
A: It is of course dangerous to say something cannot be done (and difficult to prove!), but one may be skeptical that such a (natural) problem exists.  The number of ``probable primes" that are composite (these are usually called "pseudoprimes") up to $x$ is at most 
$$ 
\ll x \exp \Big( - c \frac{\log x \log \log \log x}{\log \log x}\Big),
$$ 
by a result of Pomerance. Thus the set of probable primes is essentially just the set of primes together with a much smaller set of pseudoprimes, and I don't think we understand enough about this small set to use it meaningfully (in contrast for example with almost primes which are at least as numerous as the primes and have a more tractable distribution).  
A: There are infinitely many Mersenne probable primes.  Let $p$ be a prime (or even a probable prime).  Then I claim that $2^p-1$ is a probable prime.
Proof: $2^k \equiv 1 \pmod{2^p-1}$ if and only if $k \equiv 0 \pmod p$, so it suffices to check that $2^p-1-1 \equiv 0 \pmod p$, and this follows from $p$ being a probable prime.
A: There are infinitely many probable primes of the form $(n-1)^2+1$.
This generalizes to more polynomial since Fermat numbers satisfy
$F_n=(F_{n-1}-1)^2+1$.
