Are we supposed to count the rectangles?

For $1\leq a,b \leq 6$ , the number of $(7-a)\times(7-b)$ rectangles is $ab$ . Therefore, the total number of rectangles is $\sum_{a=1}^6\sum_{b=1}^6 ab = \left(\sum_{a=1}^6a\right)\left(\sum_{b=1}^6b\right) = (21)(21) = 441$

From this, we subtract the number of $1\times 1$ , $2\times 1$ and $1\times 2$ rectangles, to give $441 - (6)(6) - (5)(6) - (6)(5) = \boxed{345}$

Each pair of points not on the same row/column defines a unique rectangle (as they form one of its diagonal.)

So for each $(6+1)(6+1)=49$ points of the grids, we have $49 \times (49-13)$ rectangles (as there are $6+6+1 = 13$ points on the same row and columnas a given point.)

Note that a same rectangle $ABCD$ will be obtained from $4$ different pairs of points ( $(A,C),(C,A),(B,D),(D,B)$ ), for a total of $\frac{49(49-13)}{4}$ rectangles in the grid.

There are $6 \times 6 = 36$ $1$ -cell rectangles and $2 \times 5 \times 6 = 60$ $2$ -cells rectangles that we need to substract hence the answer is $\frac{49(49-13)}{4}-36-60 = \boxed{345}$