No red allowed

Let the Horizontal lines from bottom to top be $h_{0}, h_{1}, ..., h_{6}$ and Vertical lines from left to right be $v_{0}, v_{1}, ..., v_{6}$

Total Rectangles $= 7C_{2} \times 7C_{2} = 441$

Rectangles that contain bottom left red square is all rectangles that can be formed by choosing one horizontal line from $\{h_{0}, h_{1}\}$ , another horizontal line from $\{h_{2}, ..., h_{6}\}$ and one vertical line from $\{v_{0}, v_{1}\}$ , another vertical line from $\{v_{2}, ..., v_{6}\}$ .

Total number of rectangles that contain bottom left red square $=$ $(2C_{1} \times 5C_{1}) \times (2C_{1} \times 5C_{1}) = 100$

Similarly, total number of rectangles that contain top right red square $=$ $(2C_{1} \times 5C_{1}) \times (2C_{1} \times 5C_{1}) = 100$

Total number of rectangles that contain both the red squares are all the rectangles that can be formed by choosing one horizontal line from $\{h_{0}, h_{1}\}$ , another horizontal line from $\{h_{5}, h_{6}\}$ and one vertical line from $\{v_{0}, v_{1}\}$ , another vertical line from $\{v_{5}, v_{6}\}$ .

Total number of rectangles that contain both the red squares $=$ $(2C_{1} \times 2C_{1}) \times (2C_{1} \times 2C_{1}) = 16$

From principle of inclusion-exclusion:

Rectangles without any red squares $=$ $441 - (100+100-16) = 257$

We will use the inclusion-exclusion principle: The answer is (all rectangles) - 2(rectangles containing the red square in the top right) + (rectangles containing both squares) = $(\frac{7*6}{2})^2$ - 2 * 25 * 4 + 4 * 4 = 257. ) To understand this way of counting, draw a Venn diagram! To count all rectangles, we observe that there are $\binom{7}{2} = \frac{7*6}{2} = 21$ choices for the left and the right edges and as many for the lower and the upper edges. To count the rectangles that contain the red square in the top right, note that there are 25 choices for the lower left corner and 4 choices for the upper right corner. (The lower left corner has to be to the left and below the red square, while the upper right corner has to be to the right and above.) By symmetry, there are just as many rectangles (namely, 100) containing the red square in the lower left. Likewise, for the rectangles containing both squares there are 4 choices for the lower left corner as well as for the upper right corner.

I've simply counted all the possible rectangles with area 1(1×1),2(1×2),3(1×3),4(1×4 or 2×2)..and so on..that do not include the two red squares. It's very simple and due to the symmetry of the two red squares for example rectangles 1×3 are as many as rectangles 3×1. 34+52+40+28+17+16+8+24+14+8+4+2+8+2=257