Yes it can. This particular case amounts to finding a decomposition of $K_8-I$ (where $I$ denotes a one-factor) into triangles (i.e. $K_3$ subgraphs). Here each vertex represents a question, and each triangle represents a student (the vertices of the triangle are the questions answered by that student).
For the $m=1$ and $q=3$ case (as above, but with general $s$), the best possible can be derived from Alspach's Conjecture (which has recently been proved).
Alspach's Conjecture: Given positive integers $n$ and $m_1,m_2,\ldots,m_t$, then there exists a decomposition of either $K_n$ (if $n$ is odd) or $K_n-I$ (if $n$ is even) into cycles of length $m_1,m_2,\ldots,m_t$ if and only if
- $\sum_{i=1}^t m_i = {n \choose 2}$ if $n$ is odd,
- $\sum_{i=1}^t m_i = {n \choose 2}-\frac{n}{2}$ if $n$ is even, and
- $3 \leq m_i \leq n$ for all $1 \leq i \leq t$.
So, the least $n$ for which $K_n$ contains $s$ triangles will be $\mathrm{min}\left( \left\{\text{odd } n:{n \choose 2}-sn \geq 3\right\} \bigcup \left\{\text{even } n:{n \choose 2}-\frac{n}{2}-sn \geq 3\right\} \right).$
For general parameters, some particularly efficient cases are uniform block designs, such as Steiner systems. There's a lot of literature (and conferences) on the topic, so you're not going to get a complete answer unless you restrict the parameters in some way.