I try to find simple example and as many as I can and the only one im thinking of is arctan(x) -π/2--> π/2 and then i dont know how to make A and B from it
edit:
Let A={$\frac{m^2-n}{m^2+n^2}: n,m \in \mathbb{N}, m>n $} n and m odd
Let B={$\frac{m^2-n}{m^2+n^2}: n,m \in \mathbb{N}, m>n$}n and m even
probably good enough becuase Sup =1 for both and inf = 1/2 for both but is it has to be that hard?
edit2: no the sups arn't good
edit3: Let A={$a\in Q | 1<a<3 $}
Let B={$b\in R/Q | 1<b<3 $}
so $a{\displaystyle \neq b } $ for every a b and supA=supB=3, infA=infB=1
Let's use your idea of using the $\arctan$ function. In particular, we'll use the fact that it's a strictly increasing function with limits $\pm\frac{\pi}{2}$ at $\pm\infty$. Indeed consider the sets
$$A=\{\arctan(2k):k\in\mathbb{Z}\}$$
and
$$B=\{\arctan(2k+1):k\in\mathbb{Z}\}.$$
It may or may not be intuitively clear why these work, but let's prove it regardless.
We start by showing that $A$ and $B$ are disjoint, i.e. that they have no elements in common. Suppose, for a contradiction, that they did, i.e. that there is some $x$ such that $x\in A$ and $x\in B$. Then, as $x\in A$, we can write
$$a=\arctan(2k)$$
for some $k\in\mathbb{Z}$, but similarly, as $x\in B$, we can also write
$$a=\arctan(2m+1)$$
for some $m\in\mathbb{Z}$. But as $\arctan$ is injective, this would mean that $2k=2m+1$, which is a contradiction, as an integer cannot be both even and odd. Thus $A$ and $B$ are disjoint sets.
Let's now prove that $\sup A=\sup B=\frac{\pi}{2}$. Firstly, as $\arctan(\mathbb{R})=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, it is clear that $\frac{\pi}{2}$ is an upper bound of both $A$ and $B$. We show now that there is no smaller upper bound of $A$ (an almost identical argument can be used for $B$). Suppose for a contradiction that $A$ had a smaller upper bound $\alpha<\frac{\pi}{2}$. As $\arctan(2k)\to\frac{\pi}{2}$ as $k\to\infty$, we can find a $k\in\mathbb{Z}$ such that $\arctan(2k)>\alpha$ (this is just using the definition of the limit). But as $\arctan(2k)\in A$, this contradicts that $\alpha$ is an upper bound of $A$. Now just repeat the argument for $B$.
We have thus shown that $\sup A=\sup B=\frac{\pi}{2}$. An almost identical argument shows that $\inf A=\inf B=-\frac{\pi}{2}$.
Of course there are a lot of other example you could make of this, some of which use, for example, the density of $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ in $\mathbb{R}$, but as we've shown, those are not the only examples.