What is the difference when we measure torque/angular momentum about a point and about an axis? When do we measure torque about an axis and when do we measure torque about a point? What's the difference between measuring torque about an axis or a point.
I tried searching this on google but did not get satisfactory answer. 
 A: Everything in classical mechanics, momentum, angular momentum, torque, velocity etc. is measured about a point. Period. You can be sort of a Newtonian Nazi and complain that it is wrong to talk about torque about an axis and you'll be correct but here it means a completely different thing but in common language, we often make do with such words. 
So, coming to torque about an axis, it means the component of a torque pointing across a fixed direction which is along a hypothetical thing, we call the axis of rotation. The torque too is measured about a point lying on that axis and it can be geometrically proved that the component of torque along this particular direction is equal for all the points lying on this particular axis.
So, in short, torque along an axis, means the part of total torque that is along that axis measured w.r.t. a point on the axis itself.
A: Before you read the rest of my answer, you must know that strictly speaking, we always calculate torque about a POINT. But, there is a way to find torque about an axis, provided that axis is the axis of rotation. But, when you find torque about a point, you don't even need any rotational motion.
Since you have trouble understanding what torque about a point really is, consider these situations.
1.You are standing at the centre of a circular track, and a car revolves around you.
2.You are waiting to cross a road. Suddenly, a car whizzes past you. The car is not really rotating or revolving around you. But, in order to observe the car, you need to turn your head.
In these two situations, considering the car is moving with increasing speed, the car would have an angular momentum about the point of observation.Since I mentioned that the car moves with increasing speed, we know that friction causes the car's acceleration. Thus, we would say that the angular momentum of the car is changing due to torque of frictional force acting on the car about a point(you are the "point")

First, let me tell you how to calculate the torques, and then I will tell you when to apply which method.
To calculate the torque of a force acting on a body about any point, we perform the operation $\vec{\tau}$ = $\vec{R}$×$\vec{F}$ where,
$\vec{F}$ is the force vector, $\vec{R}$ is the radius vector that starts from your reference point and extends towards the point of application of the force.
By performing the cross product, you will get the direction of torque due to that force.But, you need to use the right hand thumb rule to get the "rotational effect" caused by the torque. If you are solving problems in mechanics, it is often easier to find the magnitude of the torque acting, by using T = Force applied * Perpendicular component of distance from the reference point to the point of application. To get the "rotational effect" caused by the torque, you can just imagine in your head how the applied force would actually cause the body to change its angular momentum.
Now, to find torque of a force about an axis, 
1.Choose any point on the axis
2.Find torque of the force about that point, using the method described above.


*After you have found the cross product, take the component of the torque vector ALONG the direction of the line.This gives you the torque of a force about an axis.


Coming to when to use which method....
For most standard mechanics problems that you may come across, you will mostly be dealing with rigid bodies like rolling spheres, cylinders, discs etc.
Let us say a sphere is rolling without slipping on the ground. Assume that it has an angular acceleration. Now, the only source for this angular acceleration would be the frictional force from the ground on the sphere. Here, you can take the torque of friction about the centre of mass of the sphere, or about an axis passing through the centre, about which it rotates. It doesn't matter, as you will get the same torque. Here, notice that the torque about the centre of mass would be as simple as calculating the torque about the axis, taking the arbitrary point on the axis as the centre itself. We don't have to again take any components of this torque.
Now, let me come to when it is convenient to find torque about a point.
Let us take the same example of the sphere rolling, but this time it is slipping on the ground also. If we choose any arbitrary, stationary point on the ground, the line of action of frictional force passes through that point. Thus, torque of friction on the sphere about that point is zero. Thus, angular momentum of the sphere about that point will remain conserved. So, we can very quickly calculate, for example,angular velocity of the sphere as a function of the velocity of its centre.
This method is actually extremely effective, as one does not even have to know the magnitude or direction of friction acting. You can just blindly conserve angular momentum, provided net external torque on the sphere is zero. I encourage you to solve the same situation using standard Newton's Laws. You may have to solve around 4 linear equations to get the same answer. Notice that here, we found torque about a point, and not about the axis of rotation
A: We   consider torque about point  in mechanics .It comes from the fundamental defination. 
Torque about an axis is taken when an body is hinged about that axis . We consider torque about it because only the component of torque that is along the axis is responsible for rotation rest gets cancelled 
A: I will copy the definition of torque from the wiki article:

Mathematically, torque is defined as the cross product of the vector by which the force's application point is offset relative to the fixed suspension point (distance vector) and the force vector, which tends to produce rotational motion.

 
From the definition it is clear that torque acts on a point as it is a cross product of vectors , at a "point of application" .
The axis is, as other answers have observed,   mathematically constrained to a one dimension series of points.Actually it is the vector r and the vector F which define a plane , and the perpendicular to the plane at the  origin point of r define always an axis.
If there is a physical rigid axis passing through the point the concept of torque is useful. 
In the usage of the torque quantity  in lever arms to screw or unscrew, and for calculating loads on machines having rotational parts, the axis is part of the system, but  it is at a point on the axis where the rotational force will appear.

When do we measure torque about an axis and when do we measure torque about a point? What's the difference between measuring torque about an axis or a point.

Thus, torque is always applied to a point, but the axis, real or imaginary is always there due to the cross product definition of torque.
A: Torque never acts over an axis it acts only at the point of contact whereas moment of inertia acts along an axis
A: Torque is equal in all the point of axis of rotation but what about other exis say if body is rotating on z-axis 60 ft lb. Whether it will be same on Y or X- axis.
A: Torque is always measured about an axis (in 3D) because all points along the line of action of a force yield the same result.
Consider a force $\vec{F}$ with direction $\hat{e}$. Now consider two points A and B along the line of the force. Lets say one is located at $\vec{r}_A$ and the other at $\vec{r}_B = \vec{r}_A + d \,\hat{e}$.
You can show that $\vec{\tau} = \vec{r}_B \times \vec{F} = \vec{r}_A \times \vec{F}$
$$ \vec{r}_B \times \vec{F} = (\vec{r}_A + d\,\hat{e}) \times \vec{F} = \vec{r}_A \times \vec{F} + d\,\hat{e} \times \vec{F} = \vec{r}_A \times \vec{F} $$
The same applies with linear velocity which can be viewed as the moment of rotational velocity $\vec{v} = \vec{r} \times \vec{\omega}$. Any point along the rotation axis (with direction $\hat{e}$) will result with the same linear velocity $$(\vec{r} + d\,\hat{e}) \times \vec{\omega} = \vec{r} \times \vec{\omega}$$
You would need to measure torque about a point, only in 2D problems. In reality the 2D point represents an axis coming out of the plane.  
