Assuming the prime tuples conjecture, all of these questions have affirmative answers. For instance, one can use the Chinese remainder theorem to find $a,b$ such that the tuple
$$ an+b, \frac{an+b+1}{2^2 \times 5}, \frac{an+b+2}{3 \times 7}, \frac{an+b+3}{2 \times 11}, an+b+4$$
(for instance) are an admissible tuple of linear forms (in that they have integer coefficients, and for each prime $p$ there exist a choice of $n$ in which all forms are simultaneously coprime to $p$); concretely, one can take $a=2^2\times 3 \times 5 \times 7 \times 11 = 4620$ and $b=19$. The tuples conjecture then shows that there are infinitely many $n$ in which these forms are all simultaneously prime, giving infinitely many gaps of order $3$ and length $4$. Similarly for other gap orders and lengths.
On the other hand, unconditionally not much can be said. I think at our current level of understanding we cannot even rule out the ridiculous scenario in which every sufficiently large prime gap $(p_n,p_{n+1})$ contains a semiprime. (Given that semiprimes are denser than the primes, it is likely that most prime gaps contain a semiprime, but it is highly unlikely (and inconsistent with the prime tuples conjecture) that all of them do.) The "bounded gaps between primes" technology of Goldston-Yildirim-Pintz, Zhang, Maynard, and Polymath does provide prime gaps of short length, but the method also inevitably produces several almost primes in the vicinity of such gaps, and with our current techniques we cannot prevent one of these almost primes being a semiprime inside the prime gap.
As noted by Will Jagy in the comments, this is closely related to the size of prime gaps: Any gap of size at least $\sqrt{m}$ has this property.
In fact, every gap with this property has size at least $\sqrt{m}/\log m$: We have $$\prod_{i=n}^{m } i\geq \prod_{\substack{p \leq \sqrt{m}\\ \textrm{prime}}} p$$ and the RHS has size $e^{\sqrt{m}}$ by the prime number theorem while the left hand size $\leq m^{m+1-n}$ so we must have $(m+1-n) \gtrapprox \frac{\sqrt{m}}{\log m}$.
So the conjecture that finitely many of these exist is stronger than the statement that there are finitely many prime gaps of size $\sqrt{m}$ but weaker than the statement that there are finitely many prime gaps of size $\leq (1-o(1))\frac{\sqrt{m}}{\log m}$. It's certainly conjecturally true as even stronger bounds are conjectured, but beyond the reach of proof.
From my point of view the remaining thing to do is to shrink the gap between the minimum gap size required for every small prime to appear and the minimum gap size which guarantees that all small primes appear.
To that end, observe that for $p\leq \sqrt{m}$ for the gap to have size $\leq m-n$, we must have $ \left\{ \frac{m}{p}\right\} \leq \frac{m-n}{p}$ so if $p_1 \leq p_2$, the distance from $\frac{m}{p_1} - \frac{m}{p_2} = \frac{m (p_1-p_2)}{p_1p_2} $ to the nearest integer is at most $\frac{m-n}{p_1}$.
The results of Zhang and Maynard on bounded gaps between primes count pairs of primes with bounded gaps in dyadic intervals. I think the same methods without modification count pairs of primes with bounded gaps in slightly smaller intervals, so we can choose $p_1= (1+o(1)) \sqrt{ \frac{m}{\sqrt{2}}} $ so that $p_2-p_1=O(1)$. Then $\frac{ m(p_1-p_2)}{p_1p_2}$ will be $1+o(1)$ times a bounded integer times $\sqrt{2}$, so its distance from the nearest integer will be bounded away from $0$ by an absolute constant, so $\frac{m-n}{p_1}$ must be greater than an absolute constant, and thus the gap must be greater than an absolute constant times $\sqrt{m}$.
Best Answer
oeis.org/A209407 is the smaller prime in those gaps, and oeis.org/A277718 is the larger prime. The gaps themselves do not appear to be on OEIS.