Why normal reaction, tension, and friction forces are reference frame-independent? I can not provide a whole manuscript of examples where I can justify that these forces are reference independent.
I don't think that there is any such case where these forces are reference-dependent. What is the main reason that they are reference independent?
Please provide the solutions which are in the domain of Newtonian mechanics. Every suggestion is important to me.
EDIT
A pure mathematical description of these forces would be quite helpful for me instead of specific cases.
 A: As far as I know, in Newtonian mechanics forces don't have an independent definition apart from what's given in Newton's laws. The best you can get is an intuitive understanding of a “push” or a “pull”.
Taken from Feynman's Lectures on Physics, Volume 1, (http://www.feynmanlectures.caltech.edu/I_12.html)

If we have discovered a fundamental law, which asserts that the force is equal to the mass times the acceleration, and then define the force to be the mass times the acceleration, we have found out nothing. We could also define force to mean that a moving object with no force acting on it continues to move with constant velocity in a straight line. If we then observe an object not moving in a straight line with a constant velocity, we might say that there is a force on it. Now such things certainly cannot be the content of physics, because they are definitions going in a circle. The Newtonian statement above, however, seems to be a most precise definition of force, and one that appeals to the mathematician; nevertheless, it is completely useless, because no prediction whatsoever can be made from a definition.
The student may object, “I do not like this imprecision, I should like to have everything defined exactly; in fact, it says in some books that any science is an exact subject, in which everything is defined.” If you insist upon a precise definition of force, you will never get it! First, because Newton’s Second Law is not exact, and second, because in order to understand physical laws you must understand that they are all some kind of approximation.

Even when you study a system from an accelerating frame of reference, the amount of “push” still remains the same because you're only changing the frame of reference, not the nature of the force.
Since Newtonian mechanics only deals with forces in an inertial frame (if you're analyzing things in  non inertial frames, you end up using pseudo forces, which is just a mathematical trick to convert the frame into inertial), you're never really dealing with the actual “values” of forces in non inertial frames.
A: It's because mass and time are the same in all inertial frames in Newtonian mechanics. If you have two inertial frames, the change in the velocity of an object in a given time interval would be the same as measured from both frames. Sure the actual velocity of the object measured may be different by some incremental factor but the change is measured the same. For instance, if you have two frames with a relative velocity of 2 m/s between them, then if one measures the velocity of an object at an instant as 5 m/s then the other measures it as 7 m/s at the same instant (it could be 5 and 3 too, depending on the direction of the relative motion but that's just a minor detail with no consequences here), and if at some other instant its 9 m/s for one then for the other one it would be 11 m/s, always such that the difference in the measured value is always 2. This means that if the velocity increases by 1 m/s in 1 s in one frame then it must be also true in the other frame for the difference in measured velocities to still be 2. Since the change in velocity is measured the same, the acceleration of the object is the same in both frames. Since both mass and acceleration are the same in both frames, their product, i.e, the force is also the same. This is true for any real force, not just tension, friction and normal reaction forces.
Things are different for non-inertial frames though, but the value of the actual forces like tension and stuff still stay unchanged. What you would rather see is an additional pseudo force causing the apparent extra acceleration of objects. The pseudo force doesn't get added to the real forces like tension and stuff, its seen as a completely separate force, but it's not a real force in the sense that it changes or completely disappears in some cases under reference frame changes. Unlike actual forces like friction and tension whose value doesn't change at all.
A: Newton's second law  $F=ma$ already assumes that forces are independent of the frame of reference, as far as the frame of reference is inertial. This is because the acceleration itself is invariant ($a_r=\frac{ d(v-v_r)}{dt}=a$, if $v_r$, the velocity of the new frame of reference, is constant), and the mass is assumed to be invariant too. If the new frame of reference is accelerated you still can assume that the forces are the same, but you need to introduce pseudo-forces.
In a comment it is stated that the Lorentz force is not invariant. At low speeds, which is the realm of Newtonian mechanics, it is invariant (see here). Only the complete Lorentz force is invariant, not the individual components.
In a relativistic framework forces are not constant because of the transformation of velocities makes the acceleration not invariant. In addition, the coordinate acceleration cannot increase forever, so if we define force as $F=ma$, a force cannot remain constant even within a given reference system.
A: 
Let’s look at this example.
Your choice of the generalized coordinate can be q1 or q2 or q3. To calculate the equation of motion you give the components of the position vector R in inertial space, which can be coordinate system 1 or 2 or 3.
So the equation of motion is a function of g1 or q2 or q3.
But to calculate the normal force N , you allows a deflection towards the force N (qn), which don’t depend of a coordinate system and is unique.
Fazit 
You have different choices to choose the generalized coordinate and your inertial systems, but only one choice of coordinate , which is independent of a coordinate system to calculate the reaction force.
For a conservative system , the equation of motion don’t depend on the reaction force.
I am not sure if this answer your question?
A: Simple reasoning is that we generally consider a reference frame only in manifestating the motion according to that frame of reference. For forces these reference frame does not mean any thing. It might be any force. Only the motion that comes as a resultant of the force can be considered in a reference frame. Now coming to your question. But when you consider ordinary forces they are frame dependant because the acceleration is frame dependant. When you consider frictional force they are manifestations of gravity. They remain the same what ever might be the case. Similarly tension which is inertia in the rope. 
A: All the forces you have mentioned here are manifestations of electromagnetic force, with very negligible effect from other fundamental forces. Now if we look at the form of electric and magnetic forces $$\mathbf E = \frac{1}{4\pi\epsilon_0} \frac{q}{ r^2} \mathbf{\hat r}$$ $$\mathbf B = \frac{\mu_0}{4\pi} \frac{\mathbf I \times \mathbf dl}{r^2}$$ (These are for point charge and steady line current, look up the general forms for more general cases.) You can see that the forces only depend upon quantities that are independent of frame of reference. $r$ here is the distance between the point charge(Or steady line current element) and the point at which we want to find the field. $r$ therefore only depend upon difference in position, and as mentioned in an earlier answer change in frame doesn't alter these 'difference' quantities.
