a) Yes, for example $ \{1,5,7,11\}$, since $ 1+5+7=13$, $ 1+5+11=17$, $ 1+7+11=19$ and $ 5+7+11=23$. b) No, because if there are three of them that are equal mod 3, they sum to $ 0\bmod{3}$, so that sum can't be a prime. Also, we can't have three that are pairwise distinct mod 3, since $ 0+1+2\equiv 0\bmod{3}$. So by pigeonhole we can have at most $ 2\times2=4$ numbers, as desired.

Again, the general form of b) is: In any $ 2n - 1$ integers, we can find $ n$ such that their sum is divisible by $ n$. Search for Erdös-Ginzburg-Ziv if you want.

a) yes, for example $\{3,3,5,5\}$

oh ,sorry I don't see that there is the positive integers are distnic!

hi, a)yes, the examples are in file!

I try to upload a file but can't it says "the extension out is not allowed", example is $\{1,3,7,9\}$

a) example: $3,7,9,19$ b) if look at remainders modulo 3. If we have $0,1,2$, we have a number divisible by 3 which can't be prime. So we have only $2$ remainders(if we have only one, then every sum of 3 numbers is multiple of 3). Using pigeonhole's principle we have 3 numbers given the samen remainder modulo 3, so their sum is multiple of 3, so it is not prime($1+1+1=3$ is the only such pair, but they are not distinct)

for a,it suffices to consider 1,3,7,9.

moreover,do there exist infinite number of pairs of such quadruples?

Related:What is the maximal $n$ such that there are $n$ integers (possibly negative) satisfying that condition(the sum of every three numbers is a prime number)?