a) The set ${2019!}^{2019!} ,{2019!}^{2\cdot 2019!} ,\dots {2019!}^{2019\cdot 2019!} $ works. b) Take $p$ any prime and let $a_1<a_2,<\dots < a_n$ , we have $a_2+a_3+\dots +a_{p+1} \equiv 0 \pmod p$ and $a_1+a_3+\dots a_{p+1} \equiv 0 \pmod p$ so $a_1 \equiv a_2 \pmod p $ for every prime wich means $a_1=a_2$ . So no such set exists.

For b) I had the exact same solution as @above But for a) I had a slightly different construction and would provide the motivation here. Basically we use the greedy algorithm. To counter the arithmetic mean condition we take the numbers to be multiples of $2019!$ so that when we sum any set of numbers we can factor out $2019!$ and this will be divisible by any denominator in the range. Then the geometric mean condition makes us look at the exponents which we can counter that easily just by putting a $2019!$ in the exponent as well and we are done.

Here’s my construction. $$\{2019!\cdot 1^{2019!},2019!\cdot 2^{2019!},\dots,2019!\cdot 2019^{2019!}\}$$