The sum of the areas of two squares is 640 m2.

Solution:

Let the length of the side of the first and the second square be $x$ and $y$, respectively.

According to the question:

$x^{2}+y^{2}=640 \quad \ldots$ (i)

Also,

$4 x-4 y=64$

$\Rightarrow x-y=16$

$\Rightarrow x=16+y$

Putting the value of $x$ in (i), we get:

$x^{2}+y^{2}=640$

$\Rightarrow(16+y)^{2}+y^{2}=640$

$\Rightarrow 256+32 y+y^{2}+y^{2}=640$

$\Rightarrow 2 y^{2}+32 y-384=0$

$\Rightarrow y^{2}+16 y-192=0$

$\Rightarrow y^{2}+(24-8) y-192=0$

$\Rightarrow y^{2}+24 y-8 y-192=0$

$\Rightarrow y(y+24)-8(y+24)=0$

$\Rightarrow(y+24)(y-8)=0$

$\Rightarrow y=-24$ or $y=8$

$\therefore y=8 \quad(\because$ Side cannot be negative)

$\therefore x=16+y=16+8=24 \mathrm{~m}$

Thus, the sides of the squares are $8 \mathrm{~m}$ and $24 \mathrm{~m}$.