= Solution
Put $J=\{k\in\mathbb N:k\le100\}$, avoiding any dependence on whether the convention includes $0$ among the <natural numbers>. Under <set inclusion>, the results are
|| Family
|| Maximum
|| <Least upper bound> in $A$
| $S_1$
| None
| None
| $S_2$
| $J$
| $J$
| $S_1\cap S_2$
| None
| $J$
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.
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