347. Find an arithmetic progression of five prime numbers: a) with a difference of 6, b) with a difference of 12.
347. The first term of the progression cannot be 2 and 3, because then the other terms would not be prime. The first term can be of the form $6k+1$ or $6k+5$. Let's consider these cases.
1) $a=6k+1$, such numbers can end in $1,3,5,7,9$. According to the condition, the terms of the progression must be prime numbers, so let's consider numbers ending in $1,3,7,9$ (they cannot end in 5, as they would not be prime). If a number ends in 1, the last term will end in 5 and will not be prime; if it ends in 3, the third term will not be prime; if it ends in 7, the fourth term will not be prime; if it ends in 9, the second term will not be prime.

Therefore, the first term cannot be of the form $a=6k+1$.
2) $a=6k+5$, for $k=0$ we have the progression of prime numbers $5,11,17,23,29$. In the general case, numbers of the form $6k+5$ can end in $1,7,3,9$. Such numbers were considered above.

Thus, there is a unique arithmetic progression of 5 prime numbers with a difference of $6: 5,11,17,23,29$.

Reasoning similarly, we find the unique arithmetic progression of five prime numbers with a difference of $12: 5,17,29,41,53$.