As you say for part a, if you have a group of 5 people, there has to be some lock they aren't able to open... and this lock has to be different for each group of 5 people, as if there were two different groups of 5 people that couldn't open the same lock, then their union (which would have to have at least 6 people) wouldn't be able to open the lock... but every group of 6 needs to be able to open the safe. So there needs to be at least one lock for each group of 5 people.
Now, similarly, for each group of 5 people, we have a unique lock that that group can't open. Every other person has to have a key to that lock... so that any group of 6 that included that original 5 would be able to open that lock. So every one of those C(11,5) locks must have 6 keys, so there's at least 6*C(11,5) keys in total. These will be evenly distributed over the scientists (there's one for each group of 5 people, given to everyone not in that group, and each scientist is in the same number of groups) so the number of keys per scientist is 6*C(11,5)/11. With a bit of manipulation of the factorial definition of C(n,r), we can show that's equal to C(10,5) (indeed, in general, r*C(n,r-1)/n = C(n-1,r-1), which is exactly the result you'd get for n total scientists, where a group of r can open the safe).
Alternatively (and this is the way that Syrin was explaining it): as I just mentioned, each scientist has a key for each 5-person group they're not a member of, as that's the key they'd be needed for if they were to join that group, bringing the count up to 6. And the key that each of those groups of 5 is missing must be different for each group. So each person must hold as many keys as there are groups of 5 in the remaining 10 people... that is, C(10,5).
While no one overhear you quickly tell me not cow cow.
but how about watch phone?