How big is the role of black hole spin in forming its mass? Excuse me if i'm saying something weird, but as far as i know moving faster means increasing mass according to relativity theory, right? If you make some star spin really really fast throwing something into it at high speed and appropriate angle will it turn it into black hole? If yes, how much material it would take to make Sun a black hole?
 A: Although increasing an object's spin will increase its total energy, it will not increase its tendency to collapse to a black hole. The physical reason for this, is that conservation of angular momentum will work against any possible collapse. From a slightly different perspective, it can also be seen from the fact that the event horizon radius of a spinning black hole is smaller than that of a non-spinning black hole of the same mass.
A practical and astrophysically relevant consequence of this is that rotating neutron stars can be more massive than non-rotating neutron stars. This allows for the reverse to the OPs scenario to happen: 

Suppose a neutron star is created with a very high spin (maybe from the collision of two other neutron stars), and a mass that is higher than the critical mass for a non-rotating neutron star to collapse to a black hole. Initial the neutron star is kept stable by its angular momentum, but over time it will lose angular momentum (e.g. due to emission of EM radiation) and spin down. At some point the angular momentum will be insufficient to prevent collapse, and the neutron star collapses to a black hole.

This leaves the question what portion of a rotating black hole's mass can be thought off as consisting of "rotational energy". This is not straightforward to answer since in general relativity there is no clear cut separation of different kinds of energy. However, some indication can be gleamed from looking at the rotational energy of neutron stars at the critical point of collapse. Table II of arXiv:1905.03656 gives values for the mass ($M$), angular momentum ($J$), and rotational energy ($T$) of such neutron stars depending on the model for the equation of state for the neutron star. For one such model these values are
\begin{align}
M & = 2.57 M_{\odot} \\
J &= 4.183\times10^{49} \text{erg s}\\
T &= 2.415 \times 10^{53} \text{erg}
\end{align}
This translates to a spin parameter
$$\chi = \frac{c J}{GM^2} = 0.719,$$
i.e. it would collapse to a black hole spinning at 72% of its maximum rate. However, the fraction of its total energy in rotational energy ($T/(Mc^2)$) is only about 5 percent. 
A: This question is not quite phrased in the way that a specialist would phrase it, but it still sort of makes sense, and basically the answer is that for a typical astrophysical black hole, the spin's contribution to the mass is pretty big.
Although a black hole has spin, the standard models of black holes that we study are vacuum solutions. There is no matter anywhere. So the spin of the black hole is actually an angular momentum that exists because of properties of empty space, which is kind of strange. However, if you want to get more concrete, you can equate this to the angular momentum of the infalling material that formed the astrophysical black hole that we're modeling.
This infalling material was ultrarelativistic, so its kinetic energy was large compared to its mass. Some fraction of this energy was locked up in transverse rather than radial motion. The fraction is fairly big, and the way we see this in the final object is that astrophysical black holes often have spins that are pretty close to the maximum spin that a black hole can have for that mass.
