Can you see external light approaching a black hole when you are within it? We can't see a BH because light can't escape from it, but when you are within a BH (assuming you would stay alive) could you see any light from extrenal stars? If so, would the ligth be very bright or just very dim?
Is it so that when light enter it will be likely that is gets a blue-shift because of the bigger gravity or would it be more red-shifted because light is 'torn apart' like all things getting on the horizon would get very long?
It seems also that space is falling faster than light within the horizon, so that would develop more red-shift?
And in both cases wouldn't the shifts be too big to maintain the light visible?
 A: Yes, an observer inside the (Schwarzschild) black hole region may receive signals from the external region if the BH is sufficiently large. The motion of an internal observer is arbitrary, it is pictured by a generic timelike curve. The only constraint is that, within a finite interval of proper time, the curve reaches the internal singularity (roughly speaking, this interval is large if the radius of the black hole is large). Within  this interval of time the unfortunate observer can be reached by particles entering the BH region and similarly directed towards the common future internal singularity.
Regarding your second question the answer is not so easy because there is a fundamental difference between the internal and the external region. 
In the external region there is a timelike Killing vector field giving rise to a preferred notion of Killing time, which is noting but the (external) temporal Schwarzschild coordinate $t$. 
Far away from the BH horizon, the spacetime becomes flat (Minkowskian) and that Killing vector approaches a Minkowskian time vector. 
When we speak about red/blue-shift phenomenon, we always imagine a pair of observers, both in the external region, whose stories are represented by timelike curves tangent to this Killing vector. 
One observer is close to the horizon and the other lives in the far flat region. 
The frequency of a EM signal emitted by the observer in the  flat region and entering the horizon or emitted by the observer close to the horizon and received by the Minkowskian observer does not change if measured using the Killing time. This is because the equation of EM field are invariant under the Killing symmetry.
However, the values of the frequency dramatically change when employing the proper time $\tau$ of the observers and the red/blue shift phenomenon takes place. The use of proper time is the appropriate one: It means that we describe physical phenomena referring to ideal clocks at rest with the relevant observers (with this notion of time the velocity of light is constant).
For our pair of  observers, the relation between interval of  Killing time $dt$ and proper time $d\tau$ is $$ d\tau = \sqrt{-g_{00}} dt\tag{1}\:.$$
(1) is an obvious consequence of the fact that the stories of the observers have no components along the spatial Schwarzschild coordinates $r, \theta, \varphi$ since they are curves tangent to $\partial_t$.
So 
$$\frac{d\tau_{Horizon}}{\sqrt{-g_{00}(Horizon)}}= \frac{d\tau_{Far\: Minkowski}}{\sqrt{-g_{00}(Far\: Minkowski)}}$$
that is, if $\nu$ is the measured frequency,
$$\nu_{Horizon}\sqrt{-g_{00}(Horizon)}=\nu_{Far\: Minkowski}
\sqrt{-g_{00}(Far\: Minkowski)} $$
All that holds true just because both the observers are stationary with respect to the Killing vector $\partial_t$ and therefore (1) is valid.
Another way to achieve the same result is noticing that, in view of the Killing equation for $\partial_t$, the scalar product $g(\partial_t, P)$ is conserved along the null-geodesic representing the particle of light, where $P$ is the four-momentum of the particle (parallely transported along the curve), whose temporal component is just proportional to the Killing-time frequency and thus constant as said.
The situation you consider is different because the internal observer cannot be stationary with respect to $\partial_t$ as it becomes spacelike therein!
There is no timelike Killing vector field at all in the internal region (otherwise the common future singularity could not exist).
The conclusion is that the observed frequency of the received signal depends on the motion of the internal observer according with general laws of GR on the subject: It can be arbitrarily red shifted or blue shifted depending on the four-velocity of the internal observer.
