1. **State the problem:** Solve the inequality $$\left|\left(\frac{2}{5}\right)^x - \left(\frac{5}{2}\right)^{-2x}\right| < 2.$$\n\n2. **Rewrite the terms:** Note that $$\left(\frac{5}{2}\right)^{-2x} = \left(\frac{2}{5}\right)^{2x}$$ because $$a^{-b} = \frac{1}{a^b}$$ and $$\left(\frac{5}{2}\right)^{-2x} = \left(\frac{2}{5}\right)^{2x}.$$\n\n3. **Substitute:** Let $$y = \left(\frac{2}{5}\right)^x > 0$$ since the base is positive. Then the expression inside the absolute value becomes $$y - y^2.$$\n\n4. **Rewrite inequality:** We want to solve $$|y - y^2| < 2.$$\n\n5. **Analyze the absolute value:** This means $$-2 < y - y^2 < 2.$$\n\n6. **Rewrite inequalities:**\n- For the left inequality: $$y - y^2 > -2 \implies -y^2 + y + 2 > 0 \implies y^2 - y - 2 < 0.$$\n- For the right inequality: $$y - y^2 < 2 \implies -y^2 + y - 2 < 0 \implies y^2 - y + 2 > 0.$$\n\n7. **Solve the quadratic inequalities:**\n- $$y^2 - y - 2 < 0$$ factors as $$(y - 2)(y + 1) < 0.$$\nThe roots are $$y = 2$$ and $$y = -1.$$ Since $$y > 0,$$ the relevant interval is $$0 < y < 2.$$\n- $$y^2 - y + 2 > 0$$ has discriminant $$\Delta = (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 < 0,$$ so the quadratic is always positive for all real $$y.$$\n\n8. **Combine results:** The inequality reduces to $$0 < y < 2.$$\n\n9. **Recall substitution:** $$y = \left(\frac{2}{5}\right)^x.$$ Since $$\left(\frac{2}{5}\right)^x > 0$$ for all real $$x,$$ the inequality $$0 < \left(\frac{2}{5}\right)^x < 2$$ holds for all real $$x$$ because $$\left(\frac{2}{5}\right)^x$$ is always positive and less than 2 for all real $$x$$ (since $$\frac{2}{5} < 1,$$ powers are positive and bounded above by 1 at $$x=0$$ and grow without bound only if base > 1, which is not the case).\n\n10. **Check endpoints:** The absolute value is strictly less than 2, so no equality cases.\n\n**Final answer:** The solution set is all real numbers $$x \in \mathbb{R}.$$