Some four clubs have exactly $1$ student in common

I am not completely sure my argument is correct, but I would say that it is not possible to have at least 3 students in common with all other groups, and at most one between 5 groups.

If we look at the size $s$ of a club, there are ${s\choose 3}$ combinations of 3 members. Any other club must contain at least one of those $s\choose 3$ combinations. Thus, the average number of times such a combination occurs among other clubs, is $449/{s\choose 3}$. If this number is strictly greater than 3, there must be at least one of those combinations that occurs among 4 other clubs. This contradicts the second assumption of no five clubs sharing more than one member. Therefore, $s$ satisfies $449/{s\choose 3}>3$. This gives that $s\geq 11$.

This means that the average number of clubs that a person participates in is at least $11\cdot 450/100=49,5$. Therefore, there is at least one person $p$ participating in at least 50 clubs. Consider these 50 clubs. By the second assumption, no five of these clubs share any other member than $p$. However, the total number of other members of these clubs is at least $10\cdot 50=500$, and there are 99 other persons. Therefore, at least one of these 99 other persons must occur in 5 of those clubs. This gives a contradiction.