Thanks for that. Alas, Shimura also has the formula for $\theta_3$ but not for $\theta_2$ or $\theta_4$. Maybe one can deduce the formulae from his more general considerations but he doesn't give the sort of explicit formulae I want. :-(

Another example: the embedding $\mathbb{R}\to\mathbb{R}^2$ taking $x$ to $(x,0)$ isn't Lebesgue-Lebesgue measurable in Nicolo's sense.

Continuity does not imply measurability in Nicolo's strong sense. There are continuous bijections mapping sets of positive measure to sets of zero measure (e.g. Cantor sets). Each subset of a zero-measure Cantor set has Lebesgue measure zero but its inverse image need not be measurable. Decomposing the subtraction map as a composite of $(x,y)\mapsto(x-y,x)$ and $(x,y)\mapsto x$ does it cleanly enough for me. The usual definition of a Lebesgue measurable function requires the inverse image of a Borel set to be Lebesgue integrable: this is weaker than Nicolo's condition.

There's a nice proof of the turning tangents theorem in do Carmo's Differential Geometry of Curves and Surfaces.

Yes, and when $p \equiv 3$ (mod 4) there are $a\in\mathbb{F}_{p^2}$ such that the Frobenius of the elliptic curve $y^2=x^3+ax$ has $F^2=-p^2$ (and so the curve has $p^2+1$ points over $\mathbb{F}_{p^2}$).

I now think Lang is wrong, since $E$ could have an automorphism of order $3$, $4$ or $6$. But then $E$ has $j$-invariant $0$ or $1728$ and so has a model over $\mathbb{F}_p$. Over $\mathbb{F}_{p^2}$ this model will have $(p+1)^2$ points.

The proof is in Lang's Elliptic Functions. Briefly speaking, given a supersingular $E$ defined over $\mathbb{F}_{p^2}$ the Frobenius endomorphism and $p$ are both totally inseparable and so differ by a factor of an automorphism. If $p\ge 5$ the only automorphism is $\pm1$ (according to Lang who doesn't quite convince we :-( ). So the Frobenius is $+p$ or $-p$ and that of the quadratic twist is the other.