One answer to p!+1 divisible by next biggest prime
to p: 9769626895044196025426395684657

Three answers to p!-1 divisible by next biggest
prime to p:

p=7841, 594278556271608991,
4259842839142238791410741595983041626644087433

Methodology:

Suppose that p'=p+2k is the next biggest prime to
p, and that (p!+1) is divisible by (p+2k).

By Wilson's Theorem, (p'-1)! is equivalent to -1
mod p', as is p!, by assumption.

Therefore (p'-1)(p'-2)...(p+1) is equivalent to 1
mod p'.

Therefore (-1)(-2)...(-(2k-1)) is equivalent to 1
mod p'.

Therefore p' is a factor of (2k-1)!+1.

In other words, p' is a factor of (p'-2k)!+1 iff
it is a factor of (2k-1)!+1. Therefore we need to search for prime
factors of (2k-1)!+1 that have a gap of 2k immediately preceding them.
Using
__
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math1/matha104.htm__,
we find that the only such p with a gap of at most 100 before it is
9769626895044196025426395684657.

Similarly, the other question asks for prime
factors of (2k-1)!-1 that have a gap of 2k immediately preceding them.
The only such p with a gap of at most 100 are 7841, 594278556271608991,
4259842839142238791410741595983041626644087433 from
__
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha105.htm__.