Why is borole not considered aromatic
A very basic way to do this would be to draw a diagram that shows the p-orbitals going above and below the plane.
For Borole, the bonding orbitals would look like as shown below (taken from the wikipedia page on Borole):


Natural Bonding Orbitals of Borole. Structure optimised using ORCA BP86-D3BJ and def2- TZVPP basis set. The calculated occupencies of the obitals going from left to right are $0.13$, $1.9$ and $1.9$ respectively.

[emphasis mine]
The point to be taken here is the sentence in bold, the occupany in the empty p-orbital of boron [left most diagram and first value given] is small compared to the other two π-bond orbitals. This means that there is no complete delocalisation in borole. Only the double bonds are in conjugation.
A reason for this is that the orbital of Boron is an empty p-orbital and so the total number of electrons that can potentially delocalise is $4$ and so not of the form $4n+2$. If anything, the compound would become &quot;anti-aromatic&quot; if conjugation were to happen and so that delocalisation doesn't take place.
Now, on your second question, the lone pair of nitrogen in pyridine doesn't contribute to the aromaticity since the lone pair is out of plane of the ring. So adding placing boron shouldn't affect the aromaticity.