Do these sets all contain an infinite number of prime numbers? Or, at least, provably (unconditionally), one of them?

Question

Let us define $A_1$ to be a set of all numbers of the form $p_1+q(p_1)$ where $p_1$ goes through a set of primes $\mathbb P$ and $q(p_1)$ is $1$ if $p_1$ is even prime and $2$ if $p_1$ is odd prime.

Let us define $A_2$ to be a set of all numbers of the form $p_1p_2+q(p_1,p_2)$ where $(p_1,p_2)$ goes through a set $\mathbb P \times \mathbb P= \mathbb P^2$ and $ q(p_1,p_2)$ is $1$ if at least one of numbers $p_1$ and $p_2$ is even and $2$ if both are odd.

$$...$$

More generally, let us define $A_k$ to be a set of all numbers of the form $p_1...p_k+q(p_1,...,p_k)$ where $p_1...p_k$ goes through a set $\mathbb P^k$ and $q(p_1,...,p_k)$ is $1$ if at least one of numbers $p_1,...p_k$ is even and $2$ if all are odd.

It is more than expected that for every $k \in \mathbb N$ a set $A_k$ contains an infinite number of primes , but how to prove that? Is anything known already? Are there some results in this direction?

If there is an infinite number of twin primes then $A_1$ has an infinite number of primes, so a case of $A_1$ is probably open (because converse also holds), but is anything known about other cases or some particular case? Can it be shown by some analysis that at least one of these sets must contain an infinite number of primes, maybe by some inclusion-exclusion principles?

If we denote by $\text{small}(A_k)$ the smallest member of the set $A_k$ then $(\text{small}(A_k))_{k=1,2,...}$ is just our familiar sequence of primorials.

Answer 1

An easy modification of Chen's theorem gives that $p-2$ (for primes $p\ge5$ say) is infinitely often either prime or the product of two primes. This shows that either $A_1$ or $A_2$ contains infinitely many primes.

My understanding of the general parity phenomenon is that one could prove the statement "either $A_j$ or $A_k$ contains infinitely many primes" whenever $j$ is odd and $k$ is even (in theory—see Terry Tao's comment below). But in particular, I don't believe we can yet show that any particular $A_k$ contains infinitely many primes on its own.