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@George: for $i=0,1$, if $Y_i \rightarrow Z \rightarrow X$ is a factorisation with $Z$ smooth then $Y_i \rightarrow Z$ must be an isomorphism over $X$. Since $Y_0$ and $Y_1$ are not isomorphic over $X$, this means that the two morphisms $Y_i \rightarrow X$ cannot both factor through a morphism $Z \rightarrow X$ for some smooth $Z$.
@LSpice: Correct me if I'm wrong, but my understanding is that unless someone gets a PhD from a North American institution, there is no particular reason to expect to find their thesis on MathSciNet.
Suppose $L$ and $L^\prime$ are bitangents with preimages $E_1+E_2$ and $E_1^\prime + E_2^\prime$ respectively. If the intersection point of $L$ and $L^\prime$ is not on $Q$, then each of the $E_i$ will be disjoint from one of the $E_i^\prime$. You can use this to build up a set of 7 disjoint exceptional curves fairly quickly. Lastly you need, for example, another exceptional curve that intersects precisely 2 of your 7 chosen curves. I'm not sure how efficient this is but it seems quicker than computing intersection numbers of every pair of exceptional curves.
Depending on how liberally you interpret "similar", there is also Merel's theorem: for $K$ any number field, the order of $E(K)_{\operatorname{tors}}$ is bounded by a constant depending only on the degree of $K$.