1. **State the problem:** Evaluate the expression $$\sqrt{\left[\frac{\left(\frac{2}{5} + \frac{1}{4} \cdot \frac{8}{3} - 1\right)}{\frac{2}{5}}\right]^2 + \frac{1}{3}} + \frac{1}{4}.$$ 2. **Calculate inside the parentheses first:** Calculate $$\frac{1}{4} \cdot \frac{8}{3} = \frac{1 \cdot 8}{4 \cdot 3} = \frac{8}{12} = \frac{2}{3}.$$ 3. **Sum inside the parentheses:** $$\frac{2}{5} + \frac{2}{3} - 1 = \frac{2}{5} + \frac{2}{3} - \frac{5}{5}.$$ Find common denominator 15: $$\frac{2}{5} = \frac{6}{15}, \quad \frac{2}{3} = \frac{10}{15}, \quad \frac{5}{5} = \frac{15}{15}.$$ Sum: $$\frac{6}{15} + \frac{10}{15} - \frac{15}{15} = \frac{6 + 10 - 15}{15} = \frac{1}{15}.$$ 4. **Divide by $$\frac{2}{5}$$:** $$\frac{\frac{1}{15}}{\frac{2}{5}} = \frac{1}{15} \times \frac{5}{2} = \frac{5}{30} = \frac{1}{6}.$$ 5. **Square the result:** $$\left(\frac{1}{6}\right)^2 = \frac{1}{36}.$$ 6. **Add $$\frac{1}{3}$$ inside the square root:** $$\frac{1}{36} + \frac{1}{3} = \frac{1}{36} + \frac{12}{36} = \frac{13}{36}.$$ 7. **Take the square root:** $$\sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6}.$$ 8. **Add $$\frac{1}{4}$$ outside the root:** $$\frac{\sqrt{13}}{6} + \frac{1}{4}.$$ **Final answer:** $$\boxed{\frac{\sqrt{13}}{6} + \frac{1}{4}}.$$