Can we make high acceleration safe to our bodies thanks to this gravity trick? According to Wikipedia, a human body can resist g-force of about $5 g$. It can be a greater value in some circumstances, but even as low as $2 g$ would be unpleasant after several seconds. This means our bodies set rather low limit to manned space travel. Even if we built a spaceship that could accelerate steady with $20 g$, it would be fatal to its crew.
We know that it's not the acceleration itself that is harmful; internal forces inside our bodies are. When a pilot experiences $5 g$, the force comes from their seat, through compressed skin and flesh; then the bones apply forces to every other pieces of flesh, they "feel" the flesh inertia; the heart builds pressure to accelerate blood; the skull pushes the brain, the brain pushes the skull; and so on. These forces create the impression of acceleration and may be harmful. Yet if accelerating forces were applied directly to every particle, the pilot would feel nothing. Gravity works this way. In free falling we experience zero-g even though we accelerate towards the Earth (or another body) according to a distant observer.
This observation led me to following concept:
 .-------. /==
 | engine <===   everything accelerates
 '-------' \==        to the left
   | |
  /   \
 /     \
/ large \        free falling
| mass  |     .  passenger craft,
\   M   /        small mass m
 \     /
  \   /
   | |
 .-------. /==
 | engine <===
 '-------' \==

The manned craft of mass $m$ falls freely towards unmanned large mass $M$. Engines accelerate $M$ just enough for the distance between the two masses to remain constant. There is of course some force from $m$ that pulls $M$ to the right. The engines act against this, so in fact they accelerate $M+m$. It's no surprise, no magic. Gravitational pull between $M$ and $m$ might be replaced by a rope or another structure that would connect the engines to the passenger craft.
The difference is: with the rope we have a force applied to a joint, then fuselage, its structure, seats, flesh, skeleton – all those harmful forces that can make screws break and lungs collapse. With gravity we have free fall, no tensions, no harm done.
I'm not asking about engines, energy source, tidal forces, economy etc. Let's assume we can build and power this set of crafts, even if $M$ should be of planetary scale.
Primary questions: Will passengers experience zero-g regardless of the whole set acceleration as I expect? or am I missing something? Do Newton's and Einstein's answers differ here?
Secondary questions: Has the concept already been discussed by scientists? What is its name (if any)?
 A: Primary question: This is very similar to the old question of what would happen if you fell into a black hole.  You are correct that you wouldn't feel the acceleration due to gravity per se, but you'd still need to worry about tidal forces.  These have complicated geometric dependence - they're negligible near the center of a planar mass $M$, and for a spherical mass they fall off like $1/r^3$ (or $1/M$ at the event horizon), so $M$ would have to be really big - you'd certainly need to stay outside of the event horizon if you ever wanted to leave your "spaceship".  A human couldn't survive crossing the event horizon of a solar-mass black hole, but could survive crossing that of a supermassive black hole.  So you'd need $M$ to be many, many times larger than the Sun, and you'd need your engines to be able to accelerate it to near-lightspeed in a reasonably short time.  From the rocket equation (I know, that doesn't apply in relativistic situations, but whatever), the amount of initial fuel you'd need scales exponentially with the final speed you want to attain.  So the initial mass ($M$ + fuel) would need to be exponentially larger than the mass of the Sun.  So not a practical mode of transportation.
Secondary answer: I don't think this concept is realistic enough for scientists to have bothered giving it a name yet.  You should name it the Maciorowski Technique or something!
Tertiary answer: the only detailed description of this mechanism that I'm aware of can be found here (the science fiction on this site varies considerably in hardness).
A: I think your right. Due to equivalence principle, to the free-falling body it will seem like it has no acceleration, as opposed to a body standing on ground which is equivalent to the ground pushing the body so that it will accelerate upward, and hence experience the normal force from the ground. Because the acceleration due to gravity will depend only on the distance between two objects, regardless of density or mass of the lighter object (or its components), provided the mass of the larger mass is constant. And in effect of uniform acceleration, there is no need for the components of a body (bones, flesh, whatever) to "push against one another" just to go along the acceleration of the pushing wall or ground.
Using the equation for gravitational force 
$$F = GMm/r^2 = ma$$
, with $m$ being the mass of the manned craft, $r$ being the distance of the craft and the center of the large mass $M$. It becomes
$$GM/r^2 = a$$
Then adjusting the $M$ and $r$ to find the $a$ suitable for your use. It is not necessary for M to be so large, if you could lower the distance r between the two objects.
Another problem comes with tidal forces. There is a difference in the force on the parts of the object nearer to the large mass and to parts farther from it. you could use 
$$a_1-a_2 = GM(\frac{1}{r_1^2} - \frac{1}{(r_1 + h)^2})$$
and use the condition that $$a_1 - a_2 <<g$$
so that the tidal force will be negligible. $h$ is the maximum height of a person. Choose a suitable r for that, and also adjust M for the desired acceleration. (Tip: having $a_1 - a_2 = g$ feels like having pull-ups) 
Then comes the problem of braking. when you brake, you have to decelerate, and when you decelerate, you will feel the normal force on the walls again, so make sure you decelerate slowly. and also, you can't stop your craft in the vicinity of the large mass (or on the large mass) if it has M much greater than earth. After all, the force due to the large mass is still there, even when you stop, so you'd feel a much greater force on the walls or ground than what you'd feel on earth.
That's my answer so far, not considering relativistic situations.
A: I am not versed in general relativity but I see a problem with your scheme. Let us take the ensemble of distant stars as our inertial frame. Whenever a body has acceleration $\textbf{a}$ w.r.t. inertial frame it experiences an inertial force equal to $\textbf{F}_{inertial}=-m\textbf{a}$, where $m$ is body's mass. Now suppose that the large mass $M$ in your example is stationary (engines turned off). Then body of mass $m$ with fall towards the larger mass with just that much acceleration as is required to balance inertial force by gravitational force acting on it, i.e. force balance on the body gives $\textbf{F}_{gravity}+\textbf{F}_{inertial}=\textbf{0}$. Now engines are turned on, and the large mass is given acceleration, say, $\textbf{a}'$. If $\textbf{a}'<\textbf{a}$ then smaller body will keep coming closer to the larger body, so to keep it at a constant distance we will need $\textbf{a}'=\textbf{a}$. If you accelerate the larger body further, smaller body will recede away from it. This is because whereas the larger body has engines to counteract inertial forces acting  on it when it accelerates by $\textbf{a}'$, smaller body has no such means but only the gravitational pull of the larger mass. If $\textbf{a}'>\textbf{a}$ then the body will recede away, which results in weakening of gravitational pull on it, and so on in a vicious circle, until the smaller body is (almost) completely detached from the larger body, after which it will be coasting with (almost) constant velocity.
