Let $ABC$ be an equilateral triangle of side length $1.$ For a real number $0<x<0.5,$ let $A_1$ and $A_2$ be the points on side $BC$ such that $A_1B=A_2C=x,$ and let $T_A=\triangle AA_1A_2.$ Construct triangles $T_B=\triangle BB_1B_2$ and $T_C=\triangle CC_1C_2$ similarly.

There exist positive rational numbers $b,c$ such that the region of points inside all three triangles $T_A,T_B,T_C$ is a hexagon with area $$\dfrac{8x^2-bx+c}{(2-x)(x+1)}\cdot \dfrac{\sqrt 3}{4}.$$ Find $(b,c).$
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 \( \boxed{ (8, 2) } \)