While reading Is there a graph with 99 vertices... I became curious about smaller graphs satisfying the property. According to Wikipedia, strongly regular graphs must satisfy the relation:
$$(v-k-1)\mu = k(k-\lambda-1)$$
If I computed correctly that would mean those graphs with $\lambda = 1$ and $\mu = 2$ must have $2n^2 + 1$ vertices and degree $2n$ (that would fit the $(99, 14, 1, 2)$ graph from the MO question with $n = 7$). $K_3$ trivially works, and for $(9, 4, 1, 2)$ we have the example here. The next candidate would be $(19, 6, 1, 2)$, but I couldn't find any mention of it. I'm not sure if that means that such an example doesn't exist, in which case I would be curious why, or if the existence is open.
No. Any strongly regular graph on a prime number of vertices, must be a conference graph, which means that if it has $p$ vertices then its parameters are $$ p,\ \frac{p-1}2,\ \frac{p-5}4\, \frac{p-1}4. $$ This is proved in the more recent of the books titled "Algebraic Graph Theory".
Andries Brouwer maintains a table of parameter sets of strongly regular graphs at http://www.win.tue.nl/~aeb/graphs/srg/srgtab.html; this is always a useful place to start.
For your particular case, if you compute the eigenvalues and their multiplicities, you will arrive at a contradiction; more information about how to do this in the source cited above.