Gauss Law in conducting slab 
I think the charge density should be 2E(o)ε0 via a rectangular Gaussian box oriented through slab but I am confused about the fields.Is it zero on first region and 2E(0) on second one?
 A: Think of it as the addition of three electric fields.
One is the external electric field and the other two are due to  the electric fields produced by the induced charges.

Equal numbers of positive and negative charges are induced on the surface of the conductor so the electric fields due to these charges have the same magnitude.  
$|\vec E_+| =|\vec E_-|  $
Outside the conductor these electric fields due to the induced charges are in opposite direction and so cancel each other out leaving the external electric field $\vec E_o$.
Inside the conductor the electric fields due to induced charges are in the same direction but in the opposite direction to the external electric field and add up with a magnitude equal to the magnitude of the electric field.
Thus the net electric field in the conductor is zero.  
$\vec E_+ + \vec E_-  + \vec E_o =0$
Now you can use a Gaussian cylinder across one of the surfaces to find the surface charge density of the induced charges.
A: No, no, it is E0 in the second one too! The negative chargues(electrons) in the conductor will go to the 1-2 surface and the positive(or lack of electrons) will go to the 2-3 surface, forming two chargued surfaces. This two chargued surfaces will produce an oposite electricc field that will neutralize E0 inside the conductor(you can use gauss to calculate the charge per area as E for each surface will be E0/2). But on the region 3, you have E0 and the electric field of the two chargued surfaces. As one is positive and the other one negative, the cancell each other, and so in region 3 you have E0 again.    
