Can I have a clarity on a probability(chess board) problem?

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Question is,What is the probability of selecting two squares in a chess board with one side in common?In our text book it is mentioned only for two 1*1 squares but what about remaining(like 2*2...)?

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Your book is imagining that you choose two of the $64\ 1 \times 1$ squares (I assume without replacement) and ask if they share an edge. If you want to consider $2 \times 2$ squares you have to define how you select them. One approach is to cut the board into $16 \ 2 \times 2$ disjoint squares, choose two of them and ask if they share an edge. Another is to recognize that there are $49\ 2 \times 2$ squares in a board, select two of those (with or without replacement) and ask if the two you selected share an edge. Both questions have well defined answers, which you might find if you follow the computation for $1 \times 1$ squares, but you have to define how the $2 \times 2$ squares are selected to have a unique answer.