If it's inside the event horizon, can this light escape it?
no , it cannot escape. Anything falling cannot see the ground, nor send message back, meaning in the sphere surface direction.
can you escape it by the speed of light itself?
no, it not sufficient
the light is red-shifted means that an object near the BH may emit or reflect this light which will be seen by a far observer as redshifted. The light of the inside cannot escape.
Unless we consider the Hawking radiation. Never observed, the theory seems very consistent. Some light from just beyond the event horizon might escape. A pair of particle-antiparticle is produced by the black hole's gravitational energy. One from the pair escapes and the other enters in the black hole, the balance reducing its mass.
In addition, there are differences between a ray of light doing its
way in a black hole or a mass about the speed of light in a black
It's a very difficult question because the proper time inside a black hole is a paradigm not acceptable all the other major theories : never a device would be able to measure it, nor any relevant measure information from the inside be able to leave the black hole.
Speed is the ratio distance on time. There is another issue with what we call time inside the black hole as seen from the outside. This question may explain it and an answer is provided : The arrow of time and the cosmology of a black hole interior .
But we can infer even if the result is a kind of opinion. The equation of the duration of the fall until the singularity hasn't any infinity in proper time. This equation depends mainly of the Schwarzschild radius and inversely of the mass/energy of the whole black hole. Then the marginal mass of what's falling doesn't account. Similarly , the fall duration may be computed in external time , but an infinity comes from a division by 0. In both cases, the mass/energy doesn't account.