From $k=1$ we get that each set has $2$ points. Let's choose two random sets $A_1$ and $A_2$. Then $|A_1\cup A_2|=3$, and since $|A_1|=2$ and $|A_2|=2$, this means that $A_1$ and $A_2$ have a common point. Let there be another set $A$. Similarly, $A_1$ and $A$, $A_2$ and $A$ also have a common point. Suppose that $A$ doesn't contain the common point of $A_1$ and $A_2$. Then we have $|A_1\cup A_2\cup A|=3$, which is a contradiction. Therefore $A$ contains the common point of the other sets. This means that any other set contains the common point of $A_1$ and $A_2$, meaning that all sets contain a common point, this being true for both a) and b).