 Solution

Put $J=\{k\in \mathbb{N}:k\le 100\}$, avoiding any dependence on whether the convention includes $0$ among the natural numbers. Under set inclusion, the results are

The set union of all three-element subsets of $\mathbb{N}$ is $\mathbb{N}$. An upper bound for $S_1$ under set inclusion would have to contain this entire infinite set, whereas members of $A$ are finite. Thus $S_1$ has no upper bound and hence no maximum or least upper bound.

For $S_2$, every member is a subset of the finite set $J$, and $J$ itself belongs to $S_2$, making it both maximum and least upper bound. The union of the three-element subsets of $J$ is again $J$, so $J$ is the least upper bound of $S_1\cap S_2$. It is not a member of that family, and no three-element member contains all the others, so there is no maximum.