Just for kicks, I googled the 10-digit example (6210001000) that I mentioned in the comment, and found that these n-special numbers are apparently called self-descriptive numbers.
For your part 1), it doesn't list any other examples for 7, and judging from the OEIS list, there aren't any others. (As a heads-up, there's some base conversion going on in the list; your example is listed as 389305 (which gives 3211000 in base 7))
(Addendum: This list gives the numbers as you actually want them, though there are a few less entries)
For part 2), the article lists the three values of $n$ without any (2,3,6, though I guess 1 doesn't either if you want to count it)
For part 3), the article gives a general formula for getting more examples in higher bases, but only one per value of $n$. If I find anything about generating all (either more formulas or a proof that this is all of them), I'll update the answer. The formula itself is
$$(n-4)n^{n-1} + 2n^{n-2} + n^{n-3} + n^3$$
The smallest infinitely often occurring prime gap, or
$$\liminf_{n\to\infty}\; (p_{n+1} - p_n)$$
is unknown as of now. Most likely, it is $2$, but the twin prime conjecture has not yet been settled.
Due to the work of Yitang Zhang and subsequent improvements by others (notably James Maynard and Terence Tao), we know that some prime gaps occur infinitely often. Zhang proved that gaps not larger than 70 million occur infinitely often, and the improvements lowered the bound to $246$ (perhaps there have been recent further improvements I'm not aware of).
Best Answer
Using a quick search with my computer, here’s what I found so far :
Looking only at the last digit, some power of $n$ must be congruent to $1$ modulo $10$, so $n$ must be congruent to $1,3,7$ or $9$ modulo $10$.
Looking at the last two digits, some power of $n$ must be congruent to $11$ modulo $100$, so $n$ must be congruent to $11,31,71$ or $91$ modulo $100$.
Looking at the last three digits, ome power of $n$ must be congruent to $111$ modulo $1000$, so $n$ must be congruent to one of $31, 71, 111, 191, 231, 271, 311, 391, 431, 471, 511, 591, 631, 671, 711, 791, 831, 871, 911, 991$ modulo $1000$.
Looking at the last four digits, ome power of $n$ must be congruent to $1111$ modulo $10000$, so $n$ must be congruent to one of $71, 1031, 2071, 3031, 4071, 5031, 6071, 7031, 8071, 9031$ modulo $10000$.
Do you see the pattern in those sequences ?