Let, $n \in \mathbb{N}$ and $(a_2,\dots,a_{n-1}) \in \mathbb{Z}$.
$$\alpha_k = (1+2n + a_2n(n-1) + a_3n(n-1)(n-2) + \dots + a_{n-1}(n(n-1)\dots4*3) ~ ) + kn!, ~ ~ k \in \mathbb{Z}.$$ Is it possible to find a $k \in \mathbb{Z}$ such that $\alpha_k$ is prime? Or a $k \in \mathbb{Z}$ such that $$\gcd((1+2n + a_2n(n-1) + a_3n(n-1)(n-2) + \dots + a_{n-1}) , kn!) = 1? $$
The answer is no. For example, let $n=7$, and all $a_i=3$. Then $$\alpha_k= 1+2 \cdot 7 + \mathrm{some \ stuff \ divisible \ by \ } 3 + k \cdot 7!$$ which is obviously a multiple of $3$ for every $k \in \Bbb Z$. In particular it is not prime.
As for the second question, it is equivalent to the first question. Indeed, fixed two integers $a,b$, there is a prime of the form $$a+bk$$ if and only if $(a,b)=1$. This is known as Dirichlet theorem.