1) Yes, have each element as a power of $n!$ i.e. $1^{n!}, 2^{n!}, \ldots, n^{n!}$. 2) Answer is no. Suppose that such an infinite set exists. Consider any two elements $m$ and $n$. For large $k$, $\sqrt[k]{\left(\frac{m}{n}\right)}$ must be irrational. Now if we take any other $a_1, a_2, \ldots, a_{k-1}$ in the set $(m a_1, \ldots, a_k)^{\frac{1}{k}}$ and $(n a_1, \ldots, a_k)^{\frac{1}{k}}$ cannot both be integers. We have a contradiction.

for part (a) , we can obviously construct such a set for all n let lcm(1,2,...,n)=L , then take S ={1^L ,2^L ,....,n^L }. this would do. for part (b) i think NO SUCH INFINITE SET EXISTS