Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in…

Total ways of choosing 2 squares $={}^{64}\mathrm{C}_2=32\times 63$

Now, each corners has two neighbours, and each $24$  squares across the outer edge of the chess board has $3$ neighbours, and remaining $36$ squares have four neighbours

Favorable ways $=\frac{1}{2}\left[4\left(2\right)+24\left(3\right)+36\left(4\right)\right]=\frac{1}{2}\left[224\right]=112.$Since, every pair counted twice. 

Required probability $=\frac{112}{32\times 63}=\frac{1}{18}$