I used to ponder this question some time ago but didn't come up with a satisfactory answer. The problem seems to be that until someone gives a formal treatment of the adjective "pairwise" in mathematics, it must remain a linguistic problem.
For what it's worth, here's what I've got. It's just some examples and some loose thoughts. I'll be perhaps using some non-standard notation or terminology but I think it's all going to be clear from context. I may also be talking nonsense, and I would ask everyone to point that out. I'd like to know if what I've written is too vague, silly, or just false so I can try to correct this answer or delete it.
We have "pairwise disjoint" and "disjoint" (let's assume that people actually use "disjoint" without "pairwise" in the wanted sense). A non-empty family of sets $\mathcal A\subseteq 2^X$ for some set $X$ is pairwise disjoint iff $(\forall A_1,A_2\in\mathcal A)\;\;A_1\cap A_2=\varnothing.$
$\mathcal A$ is disjoint iff $\bigcap \mathcal A=\varnothing.$
So what we have here is the operation $\cap$ which we can feed as many sets as we like. We can feed it two, and we can feed it all.
We have "pairwise coprime" and "coprime". A non-empty set of natural numbers $S$ is pairwise coprime iff $(\forall s_1,s_2\in\mathcal S)\;\;\gcd(s_1,s_2)=1.$
$S$ is coprime iff $\gcd(S)=1.$
In both cases "pairwise" is stronger. $\cap$ and $\gcd$ are meet operations in complete lattices whose respective join operations are $\cup$ and $\operatorname{lcm},$ and $\varnothing$ and $1$ are the meets of those whole lattices. In such cases "pairwise" will always be stronger! Let $L$ be a complete lattice, $\varnothing\neq A\subseteq L$ and $a_1,a_2\in A.$ Then $\bigwedge L\leq\bigwedge A\leq a_1\wedge a_2,$
And so $\bigwedge L=a_1\wedge a_2$ implies $\bigwedge L=\bigwedge A.$ That is, even one pair, let alone all pairs, of elements of $A$ having $\bigwedge L$ as a meet implies that the meet $A$ is $\bigwedge L.$
What happens when we give dual definitions for the join operations?
Let's say $\varnothing\neq\mathcal A\subseteq 2^X$ is pairwise full (made-up term) iff $(\forall A_1,A_2\in\mathcal A)\;\;A_1\cup A_2=X.$
Let's say $\mathcal A$ is full iff $\bigcup\mathcal A=X.$
Again, "pairwise" is stronger, which is not surprising. For $\operatorname{lcm}$ the pairwise condition becomes too strong to be worth considering -- the join of the whole lattice is then $0$ and no two numbers have the least common multiple equal to $0$.
One more remark is that the non-emptiness I've kept assuming is necessary for the conclusions to be true. If we drop it, the "pairwise" conditions will stop being stronger! For example $(\forall A_1,A_2\in\mathcal \varnothing)\;\;A_1\cap A_2=\varnothing$ is true, but $\bigcap\varnothing$ is equal to $X.$
Now, an example of "pairwise" being weaker. These might be again made-up conditions, but I think they're justified enough linguistically. Let's again consider the family $\mathcal A$, and say that $\mathcal A$ is pairwise intersecting iff $(\forall A_1,A_2\in\mathcal A)\;\;A_1\cap A_2\neq\varnothing.$
Let's say that $\mathcal A$ is intersecting when $\bigcap \mathcal A\neq\varnothing.$
In fact, why not try making up some strange example and see what happens? All we need is a "formula" we can feed as many variables as we want (perhaps finitely many). Unfortunately, when I tried to define what I mean by "formula" here, it started becoming much more complicated than I wanted. But see if this example works. Let's say that a finite nonempty set $S$ of the set of positive integers "pairwise adds up to eight" iff $(\forall s_1,s_2\in S)\;\;s_1+ s_2=8.$ Let's say that $S$ "adds up to eight" iff $\sum S=8.$ Now $\{8\}$ adds up to eight, but not pairwise. Is the pairwise condition weaker then? No, the set $\{4\}$ adds up to eight pairwise, but doesn't add up to eight.
This seems very complicated. And I think I simplified some things above. I should have probably been talking about multisets instead of sets. (That is for example $\mathcal A$ should have been allowed to be a multiset.) Also, I used the operations $\cap$, $\cup$, $\gcd$ and $+$, which all yield results from the same set as the operands. This surely can't be necessary. I don't think the case of independence of random variables can be considered by using such operations.