probability

\hspace{-16}\mathbf{(1)::}$Choosing $\mathbf{2}$ Square from total $\mathbf{64}$ Square is $\mathbf{\binom{64}{2}=2012}$\\\\ Now taking $\mathbf{2}$ square Horozontailly(Common Side) like $\mathbf{\square\!\square}$\\\\ Then There are $\mathbf{7}$ Possiabilities(Bcz Row has $\mathbf{8}$ boxes in Chess)\\\\ and There are $\mathbf{8}$ rows. in Chess Board\\\\ So Total no. of $\mathbf{2}$ Square each side Common is $\mathbf{=7\times 8=56}$\\\\\\

\bold{Similarly \;\; now\;\; taking\; 2 \; Square \; Vertically \; like} \begin{array}{c}\square \\ [-2mm] \square \end{array}$\\\\ Then There are $\mathbf{7}$ Possiabilities(Bcz Row has $\mathbf{8}$ boxes in Chess)\\\\ and There are $\mathbf{8}$ rows. in Chess Board\\\\ So Total no. of $\mathbf{2}$ Square each side Common is $\mathbf{=7\times 8=56}$\\\\\\ So Total favourable cases is $\mathbf{=56+56=108}$\\\\ So Required Probability $\mathbf{=\frac{108}{2012}=\frac{1}{18}}$

For (2) part of (1) I am Getting \mathbf{=\frac{7}{44}}$

(3) P(A) = P(B)

Here Green part is AUB

and white is A∩B

simply....
we know P(AUB)=P(A)+P(B) - P(A∩B)

and also P(A∩B) =P(A).P(B)

ATP...
P(A)+P(B)=2P(A).P(B)

or (√P(A)-√P(B))² =0

therefore P(A)=P(B)

Yes I am also getting 7/144

(forget to write 1 in extreme left)

for 1) 1st part i am also getting 118

for the 2nd part i am also getting 7144