Can black holes and neutron stars aquire relativistic mass due to their spin velocity? Ok, there are people that believe that black holes are just points but in that case let focus just on neutron stars... If the answer is 'yes' how much percent or promile of the total mass is to be considered relativistic? As they have circumference of around 60km and rotation up to 1000hz the speed is around 60 000 km/s so it may produce some relativistic effects.... but I am confused about the speed of rotation inside the star which could be different than the speed on the surface...and the surface is only a part of the whole star.
 A: Curiously, it is easier to answer the question thinking of black holes, precisely because their structure is way simpler. While stars have intricate internal structures as you mentioned, black holes are characterized only by their mass, charge, and angular momentum. Since we're focusing on the effects of angular momentum, I'll assume the black hole to have zero charge so that the expressions end up being a bit simpler.
A rotating black hole is described by the Kerr solution to the Einstein Field Equations of General Relativity. It can be shown that the total energy of the rotating black hole is given by (I'm quoting Wikipedia for the formula)
$$M = \frac{2 M_{\text{irr}}^2}{\sqrt{4 M_{\text{irr}}^2 - \frac{J^2 c^2}{G^2 M_{\text{irr}}^2}}},$$
where $M$ is the black hole's total mass, $J$ its angular momentum, $M_{\text{irr}}$ is its irrotational mass (i.e., the mass it would have if it had $J = 0$), $c$ is the speed of light, and $G$ is Newton's gravitational constant. It should be pointed out that this energy does induce a gravitational field, so the fact that the black hole is rotating leads to a different gravitational field than it would if the black hole was "standing still". This leads, for example, to frame-dragging effects. Furthermore, there the denominator of the expression won't vanish unless there is the presence of a naked singularity, which is often considered to be unphysical.
To find the non-relativistic limit, we must first consider the limit of a slowly rotating black hole (if it rotates too fast, notice that the rotational energy will increase the gravitational field and also contribute to the frame-dragging effects). Expanding the previous expression in powers of $J$, we find (I did the expansion on Mathematica)
$$M = M_{\text{irr}} + \frac{J^2 c^2}{8 G^2 M_{\text{irr}}^3} + \frac{3 J^4 c^4}{128 G^4 M_{\text{irr}}^7} + \mathcal{O}(J^6).$$
Notice that all of these terms blow up in the weak field ($G \to 0$) or non-special-relativistic ($c \to \infty$) limits. After giving a lot of thinking in this result and trying to do the computation in different ways, I think this might come from the fact that the existence of black holes is an intrinsically relativistic prediction and also stirs from the fact that the black hole is, in some sense, a "point". Within classical General Relativity, a black hole is not a physical body with angular momentum distributed over its pieces and so on, but rather a region of spacetime. As a consequence, it is quite common to see $J$ being referred to not as angular momentum, but as spin, since it is an intrinsic form of angular momentum.
