Would the rod translate in this case, or only purely rotate? If you have a rigid rod in space and apply an instantaneous upward force perpendicular to the very right of the surface so that it is a pure torque of 10Nm, will it only rotate or will it also translate? I say that as the line of force does not go through the center of mass, there will only be instantaneous rotation about the center of mass and no translation.
Now let's say that we also apply a pure torque of 5Nm (force going downward so that the torques vectors on both side are in same direction) to the left side. Will the rod only rotate or will it also translate? Once again, I say that the lines of force do not go through the center of mass so there is only rotation. It does not matter if the torques are "unbalanced", they simply add to be 5Nm of rotation.
Finally, with these two things in mind, why if I instead apply the forces  in the same direction upward on the rod, intuitively it is obvious the rod will translate? In this case the lines of force do not pass through the center of mass and seem to be torques as they are perpendicular to the surface. If the forces on opposite sides are balanced in the same direction then it seems it will be pure translation. But if they are unbalanced there will be a bit of rotation as well as translation?
Also, if there is rotation for a rigid body in space, will it always be about the center of mass? 
 A: In order to have pure torque you need to apply a force couple, which consists of two equal parallel and opposite forces on the rod. The application of a single force not on the center of mass will produce both translation and rotation. A single force applied to the center of mass will produce pure translation.
Hope this helps.
A: In the first case, the rod should both rotate and translate. Because there is an angular impulse as well as a linear impulse applied to the body, to conserve the total momentum, we must have the body rotating as well as translating.
In the second case as well, the rod should both rotate and translate. Assuming that both the forces are applied at the extreme ends of the rod, the magnitude of forces must differ, which means there will be a linear impulse on the rod. And obviously there is an angular impulse on the rod due the imbalanced torque.
In the third case, if both the forces are applied in such a way that the total torque is 0, then there will be no rotation and only translation. But, if the forces are applied such that we are getting a net non zero torque, then we will have rotation as well as translation.
In general, if we have a 0 net force, then we will have 0 net linear impulse, which means no translation based on conservation of linear momentum and if we have 0 net torque, then we will have 0 net angular impulse, which means no rotational motion based on conservation of angular momentum. Based on this you can work out other cases as well.
As for the last question, I don't know whether there is a proof for this, but usually bodies in isolation rotate about there centre of mass.
A: The net external force acting on the center of mass is the vector sum of all the forces acting on the system of particles. Thus in your given examples, assuming those are the only external forces acting on the bodies, the net resultant force will act on the center of mass, causing translational motion.  Since the point of application is not at the centre of mass, there will be some rotation as well. Also, since the center of mass of the body is usually chosen to be the origin or reference point, the moment of inertia about the centre of mass is minimum, so free rotation happens about the center of mass.
