The reworked addition to the second question has a positive answer. For any positive integer $n$ we can find an initial segment of the primes $p_1,p_2,...,p_m$ (where $m > 0$) and a positive number $k$ such that neither $k$ nor $k+2n$ is divisible by any of those primes, but each integer strictly between $k$ and $k+2n$ is divisible by at least one of them.
For motivation consider the special case in which $2n+1$ happens to be a prime. Then take $k=1$ and the primes to be those less than $2n+1$. Clearly $k$ and $k+2n$ are free from those small divisors, while every integer in between must be divisible by at least one of them.
Edited: Changed to make larger primes those above $2n$ (rather than above $n$)
For the general case we will use the CRT. It is simply a matter of choosing remainders for $k$ modulo the primes less than or equal to $2n$ so that neither $k$ nor $k + 2n$ is divisible, and then deploying the larger primes as needed to knock out any in-between integers not already divided by one of the smaller primes.
Obviously we want $k$ to be odd, so that $k+2n$ is odd as well. For odd primes $p_i$ up to (and if necessary) including $n$, there will always be at least one remainder for $k$ mod $p_i$ such that neither $k$ nor $k+2n$ is divisible by $p_i$. If $p_i$ divides $2n$, then this is really just a matter of choosing a nonzero remainder at $k$. If not, the fact that $p_i$ > 3 means there are only two "bad" choices for the remainder at $k$, so there's always at least one good choice.
Now consider any integers strictly between $k$ and $k+2n$ that are not already "sieved out" by those smaller primes. Then use the successive primes above $2n$ to get one-by-one a divisibility condition on those intervening numbers, which of course specifies a nonzero remainder for $k$ modulo each such prime. A crude estimate is that we need no more than $n$ primes greater than $2n$ to accomplish knockouts of all the entries unsieved by 2 (and hence by the collective set of smaller primes).
Finally determine a $k > 0$ having the specified remainders modulo the primes $p_1,...,p_m$ and we are done. None of them divide $k$ or $k+2n$, but at least one divides each of integers strictly in between.