Task 1. Given is a square board. Parallel to the sides and lines (7 lines in each direction) are drawn, which divide the board into 64 rectangular fields. These fields are colored in a checkerboard pattern. Their sizes can be different, but it is known that the ratio of the area of each white field to the area of each black field does not exceed 2. Find the maximum possible value of the ratio of the total area of the white fields to the total area of the black fields.
Solution: Let's consider the rectangles into which the board is divided by the second, fourth, and sixth vertical and horizontal lines. Each of these 16 rectangles contains four chess-colored rectangles. For each of the 16 rectangles, we will prove that the ratio of the total area of the white fields to the total area of the black fields does not exceed $\frac{5}{4}$.

Let a rectangle $a \times b$ be divided into four chess-colored rectangles, and let one of the white rectangles have dimensions $x \times y$. Then the other white rectangle has dimensions $(a-x) \times (b-y)$, and the dimensions of the black rectangles are $(a-x) \times y$ and $x \times (b-y)$. From the condition that the area of each white field does not exceed twice the area of each black field, we get

$$
\frac{a}{3} \leq x \leq \frac{2a}{3}, \quad \frac{b}{3} \leq y \leq \frac{2b}{3}
$$

Then the inequality

$$
(3x - 2a)(3y - b) + (3y - 2b)(3x - a) \leq 0
$$

holds, which is equivalent to

$$
xy + (a-x)(b-y) \leq \frac{5}{4}((a-x)y + x(b-y))
$$

i.e., the total area of the white fields does not exceed $\frac{5}{4}$ of the total area of the black fields.

Since for each of the 16 rectangles the ratio of the total area of the white fields to the total area of the black fields does not exceed $\frac{5}{4}$, this property holds for the entire board.

The value $\frac{5}{4}$ is achieved by dividing the board into 16 equal squares and for each of them choosing $x = \frac{a}{3}$ and $y = \frac{b}{3}$.