1. Identify the problem and given information:
- We have an equilateral triangle $ABC$ with side length 1.
- Points $A_1$ and $A_2$ are on side $BC$ such that $A_1B = A_2C = x$ where $0 < x < 0.5$.
- Triangles $T_A = \triangle AA_1A_2$, $T_B = \triangle BB_1B_2$, and $T_C = \triangle CC_1C_2$ are constructed similarly.
- The region inside all three triangles forms a hexagon with area given by:
$$\frac{8x^2 - bx + c}{(2-x)(x+1)} \cdot \frac{\sqrt{3}}{4}$$
- We need to find the values of $b$ and $c$.

2. **Determine the area when $x = 0$:**
- When $x = 0$, the points $A_1$ and $A_2$ coincide with $B$ and $C$ respectively, making $T_A$ degenerate and the hexagon coincides with the original triangle $ABC$.
- The area of $ABC$ is:
$$\text{Area} = \frac{\sqrt{3}}{4}$$
- Substituting $x = 0$ into the given area formula:
$$\frac{8(0)^2 - b(0) + c}{(2-0)(0+1)} \cdot \frac{\sqrt{3}}{4} = \frac{c}{2} \cdot \frac{\sqrt{3}}{4}$$
- Equating this to the area of $ABC$:
$$\frac{c}{2} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$$
$$\frac{c}{2} = 1 \implies c = 2$$

3. **Determine the area when $x = 0.5$:**
- When $x = 0.5$, the points $A_1$ and $A_2$ are at the midpoints of $BC$, making $T_A$ a smaller equilateral triangle with side length $0.5$.
- The area of each smaller triangle is:
$$\text{Area of } T_A = \frac{\sqrt{3}}{4} \left(\frac{1}{2}\right)^2 = \frac{\sqrt{3}}{16}$$
- The hexagon formed by the intersection of $T_A$, $T_B$, and $T_C$ has zero area because the triangles overlap completely.
- Substituting $x = 0.5$ into the given area formula:
$$\frac{8(0.5)^2 - b(0.5) + 2}{(2-0.5)(0.5+1)} \cdot \frac{\sqrt{3}}{4} = 0$$
$$\frac{2 - \frac{b}{2} + 2}{1.5 \cdot 1.5} \cdot \frac{\sqrt{3}}{4} = 0$$
$$\frac{4 - \frac{b}{2}}{2.25} \cdot \frac{\sqrt{3}}{4} = 0$$
$$4 - \frac{b}{2} = 0 \implies b = 8$$

4. Conclusion:
- The values of $b$ and $c$ are $8$ and $2$ respectively.

The final answer is $\boxed{ (8, 2) }$