No, it's not possible.
Why?
Think about, for a second, what you're trying to achieve. The most important concept that you're missing is that a force is an interaction between two objects. One applies a force to the other and the other applies that same force back. You might be wondering how anything gets done/moved then?
Well, if you think about the simplest mathematical expression of force $F= ma$, it shows that the force exerted is the product of mass and acceleration. When we say that an object exerts the same force on an object, it does. But the gist of it is that objects differ in mass. Sometimes drastically.
$$m_1a_1 = m_2a_2$$
$$F_{1\rightarrow2} = F_{2\rightarrow1}$$
Here's that in written form. Object 1 applies a force to Object 2. An interaction. Object 2 applies a force to Object 1. Forces are equal, yet, they move. Why? Think about it, if object 1 has $m_1 = 100m_2$ (100 times the mass of object 2) and exerts a force of, let's say, 300 N on it, it will move it without breaking a sweat. Object 2 will apply that same force on it, but because it has such low mass, it won't make even a dent in the acceleration of Object 1. That's a crucial part of that interaction. Here, even the math agrees:
$$100m_2a_1 = m_2a_2$$
$$100a_1 = a_2 $$
The acceleration of object 2 is, linearly, 100 times bigger more than object 1. Now, let's get to your problem. Can someone lift himself? You push yourself and yourself pushes you. Since we're talking about the same object, this reduces down to two same masses exerting the same force on each other. And the net force is $0$.
Same goes for cartoonish ideas like moving a boat by blowing into the sails. This is a repercussion of the translational symmetry of our Universe, conservation laws and common sense™. Hopefully, this is simple enough to make you see why it doesn't "work".
We often use the sum of torques about an axis to find the case of zero rotation by finding where the sum is zero. The reason this works is that the case of no rotation must also be a case of no rotational acceleration and therefore no change in angular momentum.
$$\tau_1 + \tau_2 + .... = I \frac{\Delta L}{t} = 0 $$
But in your case the car is accelerating. The angular momentum of your system is
$$ L = d \times v$$
The offset distance between the center of mass of the car and the axis is fixed ($0.5m$), but the velocity of the car is changing. That means the angular momentum is also changing and there is a net torque. You cannot solve for the case where net torque is zero.
In the frame of the car, the axis is no longer accelerating, but you have a fictitious force on the center of mass to account for.
Best Answer
If I understand the question correctly, the crane is firmly attached to the chassis of the truck. The bumper is assumed to be firmly attached to the chassis as well. Attaching the hook anywhere on the bumper will not lift the wheels off the ground.
The hand waving argument: This is like trying to lift yourself off the floor by using your hands to pull up on your own knees.
The formal argument would be this: The cable provides an upward force on the bumper that is exactly cancelled by the downward force at the tip of the crane. Because of vector addition, no change in angle of the crane can alter this exact cancellation. In practice you could deform the bumper, but never lift the whole truck.
If I am misunderstanding the question and the crane is attached to the ground and not the chassis, then yes any upward force would lift the front of the vehicle.
Many cranes have retractable support legs that stabilize the crane laterally. Even if the legs are engaged the crane cannot provide a torque about the rear wheels. However, the support legs often have enough travel and strength to lift the vehicle on their own. In that case, front mounted legs could lift the front of the vehicle. I've seen this done. It may even be standard procedure.