A solution (x,y) of the system  of equation $x-y=1/3$ and ${\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)=1/2$ is given by

${\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)=1/2$

$\Rightarrow \cos \pi (x+y)\cos \pi (x-y)=1/2$

$\Rightarrow \cos \pi (x+y)\cos (\pi /3)=1/2[\because x-y=1/3]$

$\begin{array}{l}\Rightarrow \cos \pi (x+y)=1\\ \Rightarrow \pi (x+y)=2n\pi \Rightarrow x+y=2n\end{array}$

Now $x+y=2n$ and $x-y=1/3$

$\Rightarrow x=n+1/6,y=n-1/6,(n\in I)$

$\therefore (x,y)=\left(n+\frac{1}{6},n-\frac{1}{6}\right)$ which is satisfied by (c) for $n = 2$