Sums of all pairs: possible less than quadratic?
Yes, it is possible using $O(n \log n)$ preprocessing and $O(\log n)$ query time.

Given the set $S$, construct the polynomial $P(x)=\Sigma_{s\in S}\textrm{ } x^s$. Then use the FFT multiplication algorithm to compute $P^2(x)=P(x)*P(x)$ in $O(n \log n)$.

A value $q$ s a sum-of-pairs of $S$ if and only if the coefficient of $x^q$ in $P^2(x)$ is non-zero. Use binary search to look up the coefficient in $O(\log n)$ time ($P^2(x)$ is a polynomial of degree $2m$).

There are also less complicated (and less efficient, but still better than $O(n^2)$) multiplication algorithms that use divide-and-conquer such as Karatsuba multiplication ($O(n^{1.585})$ preprocessing).