Potential Difference Cylinder 
In this question, I have two particular questions.
First, how do we know that Va < Vc as written in the image? The solution uses the word downhill but I don't really get what it means and how we get it.
Second, How do we know that E = 0 as noted in the image? 
Please help. Thanks.
 A: It will make more sense to answer your questions in reverse order:

How do we know that E = 0 as noted in the image?

Because the cylinder is made out of metal, we know it is conductive.
Therefore if there were an electrical field present, there would be a current flowing. But this is an electrostatics problem, so current can't flow without a complete circuit. So there must not be a current flowing. Therefore there must be no field within the conductive material.

how do we know that Va < Vc as written in the image?

Because there's no field in the metal regions, we know $V_c = V_b$.
And there's a positive charge on the outer conductor and a negative charge on the inner conductor. Therefore between the two cylinders there's an electric field pointing from the outer conductor to the inner conductor. 
And since 
$$V_{ab}=\int_a^b\vec{E}\cdot{\rm d}\vec{\ell}$$
we know that $V_{ab}<0$ so $V_a < V_b$.
A: An underlying principle in electrostatics is the the electric field within a conductor, eg a metal, is zero.
This also means that the potential is the same throughout a conductor.  
So the potentials between radii b and c are the same. 
 
Deciding which potential is the largest (or smallest) can be arrived at in many ways.  
One way is to remember that as a positive charge moves in the direction of an electric field line the potential drops.
This comes from the relationship that the electric field strength is equal to minus the potential gradient $E = - \frac{dV}{dr} $.  
In this example the charge of $-5$ on the inner cylinder induces a charge of $+5$ on the inner surface of the outer shell.
This has to happen to make sure that the net charge inside the outer shell is zero which in turn, by using Gauss's law, implies that the electric field inside the shell is zero.  
Since the outer shell has a net charge of $+2$ this means that there must be a charge of $+2-5 = -3$ on the outside of the outer shell.  
To move a positive charge from radius c to radius b no work has to be done as the conductor is an equipotential with no electric field inside it.
To move a positive charge from radius b to radius a work is done by the electric field  and so the potential at radius a is lower than the potential at radius b.  
The gravitational equivalent of this is to say that if you start at the top of a hill at radius b and slide downhill to radius a the gravitational field does work on you and so there is a decrease in the potential.  
Looking at the graph, a positive charge going from radius b to radius a is going "downhill" in the direction of the electric field.
