Electric field strength.

It would be simpler if you treat the set up as a parallel plate capacitor.

By moving plate A towards B, the capacitance of the capacitor increases, as capacitance is inversely proportional to plate separation. But capacitance = charge/potential. Given a fixed charge on the capacitor, an increase of capacitance implies a decrease of potential of plate B.

The movement of the plates doesn't affect the separation between the parallel field lines, as field lines originate from and end at the same magnitude of charges of opposite polarity. Since separation of field lines gives a qualitative measure of the field intensity, the same separation of field lines indicates the field intensity remains unchanged.
.

Note that there are induced negative charges on plate A (on the face near to B)

So when A is brought closer to B, the potential at B will decrease since the potential at B is contributed by the positive charges on itself and the induced negative charges on A. As A is brought closed, the potential contributed by negative charges on A on B will be more negative, i.e. decrease.

By E = Q/(εA) where Q is the cahrge, ε is permittivity and A is the plate area, it should remain unchanged since the charge on B is constant as it is isolated.