If gravity reaches infinite intensity on the all event horizon surface, doesn't this fact exclude the black holes central singularity? Aren't this two singularities an excess or an unnecessary redundancy?
 A: 
If gravity reaches infinite intensity on the all event horizon surface

The spacetime curvature at the horizon is finite and gravity is spacetime curvature.  Thus, gravity doesn't "reach infinite intensity" at the horizon:
Courtesy of Google books, here's the relevant section from "Gravitation":

Misner, Charles W., et al. Gravitation. Princeton University Press, 2017.
The payoff of this calculation: according to equations (31.6), none of the components of Riemann in the explorer's orthonormal frame become infinite at the gravitational radius. The tidad forces the traveler feels as he approaches $r=2M$ are finite; they do not tear him apart-at least not when the mass $M$ is sufficiently great because at $r=2M$, the typical non-zero component $R_{\dot{\alpha}\dot{\beta}\dot{\gamma}\dot{\delta}}$ of the curvature tensor is of the order $1/M^2$. The gravitational radius is a perfectly well-behaved, nonsingular region of spacetime, and nothing there can prevent the explorer from falling inward.
By contrast, deep inside the gravitational radius, at $r=0$, the traveler must encounter infinite tidal forces, independently of the route he uses to reach there. One says that"$r=0$ is a physical singularity of spacetime." To see this, one need only calculate from equation (31.4b) or (31.6) the "curvature invatiant":$$I\equiv R_{\dot{\alpha}\dot{\beta}\dot{\gamma}\dot{\delta}}R^{\dot{\alpha}\dot{\beta}\dot{\gamma}\dot{\delta}}=48M^2/r^6$$


A: A comment to the question states it is motivated by my answer to Surface gravity and mass of a black hole. Specifically that answer explains that the gravitational acceleration experienced by a user hovering at a distance $r$ from the black hole is:
$$ a=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{1} $$
and this does go to infinity as $r \to r_s$. But this does not mean that the gravity is in any sense infinite at the event horizon. All this result does is indicate how dangerous it is to draw conclusions from the experience of any particular observer. As Alfred points out in his answer, the spacetime curvature remains finite at the horizon, and indeed is finite everywhere except at the singularity (where it is undefined).
A convenient measure of the curvature is the Kretschmann scalar:
$$ K = \frac{48G^2M^2}{c^4r^6} \tag{2} $$
And this is finite for all $r \gt 0$. Alternatively another measure of the gravity at the horizon is the surface gravity:
$$ \kappa = \frac{1}{2r_s} \tag{3} $$
which again is finite.
The question is then why equation (1) gives an infinite resut while equations (2) and (3) give finite results. Speaking loosely, this is due to the time dilation experienced by an observer hovering close to the event horizon. If we label the time measured by the hovering observer by $\tau$, while the time we measure far from the black hole is $t$, then the time dilation is given by:
$$ \frac{\tau}{t} = \sqrt{1 - \frac{r_s}{r}} \tag{4} $$
As $r \to r_s$ we find the time experienced by the hovering observer $\tau \to 0$. This is the source of the (misleading) claim that time stops at the event horizon. If you'll allow me a rather hand waving argument, acceleration has units of:
$$ \textrm{acceleration} = \frac{\textrm{distance}}{\textrm{time}^2} $$
In the hovering observer's frame time $\tau$ is reduced due to the time dilation so the $1/\tau^2$ term is increased, and indeed goes to infinity at the event horizon. This is the reason the acceleration measured by the shell observer goes to infinity as the observer approaches the horizon.
The surface gravity that I mentioned in equation (3) is in effect the infinite acceleration measured by the shell observer but corrected for the time dilation, so it is a more physically realistic measurement of the gravity at the horizon. You may note that for arbitrarily large black holes this surface gravity can be arbitrarily small.
