Can an algorithm that yields $O(n^2)$ answers run in $O(n)$ time?
Can an algorithm that yields $O(n^2)$ answers run in $O(n)$ time?


Yes.  The following algorithm yields $O(n^2)$ answers (remember what $O(\cdot)$ means!) but it runs in time $O(n)$.

halt;


You probably meant to ask if an algorithm that yields $\Theta(n^2)$ answers can run in time $O(n)$. The answer to that is no. Whatever exact definition you choose for "yielding an answer", yielding one answer must take at least one computation step in any reasonable model of computation, so writing $\Theta(n^2)$ answers must take $\Omega(n^2)$ steps.