Most activated position for Electrophilic addition in substituted biphenyls I know the first structure will go fastest electrophilic addition due to +mesomeric effect but I was wondering what would be the product(major) if a electrophile attacks 

Will the electrophile attack both ring or the ring near to $\ce{OH}$ group? Also the negative charge due to resonance will resonate on both the rings then how will I decide which ring to attack to give me major product?

According to me, the ortho positions of the ring near the $\ce{OH}$ group will have a negative charge - as well as the para position - but the para position is hindered. So, the electrophile will attack the ortho positions. 

If this the correct explanation of the question, then I have one more question as well, that: when negative charge delocalises to the para position, it can delocalise to the other ring as well. Then, the electrophile could attack in the second ring also, where the negative charge would be generated by resonance. So, will the electrophile attack-the first ring or the second ring?
Dipole moment is what matters in substituted biphenyls.

The dipole moment in the first case is from left ring to right ring due to -OH groups predominant +M effect to the ring.(the alkene being an EWG(electron withdrawing group)), further enhances this effect. So effectively, electron density is more in the right hand side (I know it is counter-intuitive but it is how it is).
So EAS takes place preferably/majorly in the right side ring.

In the second case, NO2 shows -M effect and this dominates that of the alkenyl group. So dipole moment is from right to left ring and EAS takes place preferably/majorly in the left side ring.

Note: If there is any group(which is symmetric, or else we need to consider electronic effect due to assymetric grp also) between the C-C bond linking the two rings together, then the answer is as it "intuitively seems/as you expect". Try to rationalize this yourself.