We have selected 7 three-element subsets of a finite set such that any two different elements are contained in exactly one selected subset.

a) How many elements does the set have?

b) What is the maximum number of selected subsets we can choose such that no three of them contain the same element?
Let's denote the number of elements in the set by $n$. Then the number of two-element subsets of the set is $\binom{n}{2}=\frac{n(n-1)}{2}$. A selected three-element subset contains exactly $3$ two-element subsets, so $7$ three-element subsets contain a total of $7 \cdot 3=21$ different two-element subsets. Therefore, $\frac{n(n-1)}{2}=21$, from which $(n>0$ under this condition) $n=7$ follows. Thus, the set has 7 elements. (If the elements of the set are denoted by $1,2,3,4,5,6,7$, then a possible example of the seven subsets is: $\{1,2,3\} ;\{1,4,5\} ;\{1,6,7\} ;\{2,4,6\} ;\{2,5,7\} ;\{3,4,7\} ;\{3,5,6\}$. This example is also illustrated by the diagram, where the elements of the set are the points labeled with numbers, and the selected subsets are the subsets formed by the points on the drawn lines and the circle.)

We will show that we can select at most 4 subsets that meet the conditions. If we select 5 subsets, then these subsets contain a total of $5 \cdot 3=15$ elements, counting multiplicities. Since $15>2 \cdot 7$, there must be at least one element that appears (at least) three times, meaning that among the 5 subsets, there are definitely 3 that contain the same element. However, we can always select 4 subsets: fix one element of the set, denote it by 1. According to the conditions, this element is contained in exactly $\frac{7-1}{3-1}=3$ selected subsets. We will show that among the remaining $7-3=4$ selected subsets, no 3 contain the same element. If there were 3 among them that contained the same element, then the other, total $3(3-1)=6$ elements of these 3 subsets would all have to be different, so one of them would also contain 1, which is a contradiction. (In our previous example, the 4 subsets are: $\{2,4,6\} ;\{2,5,7\} ;\{3,4,7\}$ and $\{3,5,6\}$.

In summary: the set has 7 elements, and we can always select at most 4 subsets that meet the conditions.