Energy conversion of electrons in cathode rays

I haven't checked your numbers exactly but yeah your calculation sounds about right.

The energy expression in relativity is modified to $$E=gamma m c^2=frac{mc^2}{sqrt{1-beta^2}} = mc^2+frac12 mc^2beta^2+frac38 mc^2beta^4+dots$$ and of course your expression is subtracting off this first term and the second term is the classical kinetic energy, meaning the third term is the leading relativistic correction. Taking the ratio we see that it is proportional to $beta^2$ so that at $beta=0.1$ the correction to the kinetic energy is approximately 1%, and then $K+delta K=frac12 m(v+delta v)^2approxfrac12mv^2+mv~delta v$ let's us translate this error $delta K/K$ into an equivalent error $delta v/v=frac12delta K/K$, so a 1% difference in kinetic energy translates to an 0.5% difference in velocity, which is about what you are reporting.

I am not a particle physicist but I am given to understand that with these sorts of things they tend to look at the world in terms of energies rather than masses and velocities. So the 3 keV of energy just needs to be compared to the rest mass of 511 keV of energy, the total energy is 514 keV with a rest mass of 511 keV. So they would take a more direct route of saying $gamma =514/511approx 1.00587$ and computing $beta=sqrt{1-gamma^{-2}}approx 0.10788,$ whereas in this language we would have calculated the classical velocity as $beta_c=sqrt{2(gamma - 1)}approx 0.10836,$ so yeah, the error between those is indeed about a half of a percent.

In relativistic electrodynamics, the kinetic energy an electron of rest mass $m_0$ after it has been accelerated to a speed $v$ from rest, is $$KE = mc^2 -m_0 c^2$$ where $$mc^2=gamma m_0c^2$$ and the potential energy that is contained in the electric field is converted to the kinetic energy of the electron, so that $$mc^2 -m_0 c^2=qV$$

This is consistent with your question, and the result for $v$ is correct.

I guess relativistic corrections are relevant when the compared classical result deviates by $ge 1%$. The velocity ratio between the relativistic case and non-relativistic case in your example is $$frac{3.23}{3.25}approx 0.62%$$ which is close to $1%$. It also does depend on how much precision one needs, but the $ge 1%$ ratio deviation is perhaps the generally accepted figure. But I do not think there are any references stating this, as such things need not be written in stone.