I don't think that quadratic reciprocity is much faster then Euler's criterion. Both need about log p steps, which in one case consist of a product and a modular reduction, and in the other case of a modular reduction and some overhead. With some care you can avoid "if", but still you have to extract the last bits and one sign. I would guess that quadratic reciprocity overtakes Euler between 50 and 100 digits, but only if you took great care in coding reciprocity.

That depends on your application. A measure for the importance of the major arcs other than the trivial ones is the variance of the singular series. The singular series itself is usually easy to guess beforehand, so you can compare the contribution of 1 with the other major arcs. To measure the quality of the minor arc estimate compare your upper bound with an explicitly computed value. If the singular series is almost constant, and your estimate is good, you are lucky and might need only a small neighbourhood of 1. Otherwise you should include more major arcs right from the beginning.

@Joel: When passing from $\Psi$ to $\theta$, you introduce a systematic bias of magnitude $c_q\sqrt{x}$ favouring non-squares over squares. The zeros of the relevant $L$-series introduce a "random" term of the same magnitude, so when considering $\pi(x,q,a)$ we expect that most of the times the non-squares lead, but sometimes the squares overtake for a short time. But when considering $\sum\frac{1}{p}$, partial summation shows that this bias becomes $\mathcal{O}(q^{-1/2+\epsilon})$, which is negligible compared to the contribution of the small primes.

Probably the probabilistic model does not give the correct answer when dealing with primes in short intervals. Maier's matrix method shows that on a scale up to a power of $\log x$ the distribution of primes is much more irregular than what probability would predict.

$\pi$ is transcendent, hence $\mathbb{Q}(\pi)$ is isomorphic to the field $\mathbb{Q}(x)$ of rational functions. To show that the numbers $\sqrt{n^2+\pi^2}$ are $\mathbb{Q}$-linearly independent is therefore equivalent to the statement that the functions $x\mapsto\sqrt{n^2+x^2}$ are $\mathbb{Q}$-linearly independent.