finitly many primes? is it mean that the number of primes is defined? For example does sequence $ a_k = k!$ satisfy the conditions?

It's an open problem if $ k!$ works (it's not known if $ k!+1$ is prime finitely often or not). But one can take $ a_k$ as a multiple of $ k!$ such that $ a_k \pm 1$ are not prime.

OK, then we take $ a_k=k!+1$

No, as then for $ n=0$ you have no proof of finiteness of primes in $ a_k+n$.

Let $ a_k = k! + 2$, then $ a_k + n$ had only finetely prime numbers for any $ n\in N$. I am sorry $ n\in Z$. Therefore we can take \[ a_k=(k!)^6\], then $ a_k-1=((k!)^3-1)((k!)^3+1$ and $ a_k+1=((k!)^2+1)(k!^4-k!^2+1).$

May be I misunderstand the condition, but we must prove, that for every integer $ n$ the set $ {a_k+n|k=1,2,3,...}$ contains finitely number of primes. Does we vary the $ k$ or $ n$?

The set is explicitely given by holding $ n$ constant and varying $ k$.

if the $ n$ constant, then what is the problem with $ k!+1$? it's at most one prime. Or we consider unions of sets for all n?

For all $ n$ the set corresponding to $ n$ shall have only finitely many primes.

You can use also $ a_k=4(k!)^4$ and for n=1 use Sophie Germain while for n=-1 use difference of squares

Take ak=(k!)^3