1. **State the problem:** Simplify the expression $$\left(\frac{5}{2}+\frac{1}{4}-\frac{7}{8}\right) : \left(-\frac{3}{16}\right) - \frac{5}{3} \left(-\frac{2}{7}\right) + \left(-\frac{1}{2}\right) \left(-2\right)^3 - \left(3 - \frac{9}{5}\right) : \left(-\frac{1}{5}\right)$$. 2. **Simplify inside the first parentheses:** Find common denominator for $\frac{5}{2}$, $\frac{1}{4}$, and $\frac{7}{8}$. $$\frac{5}{2} = \frac{20}{8}, \quad \frac{1}{4} = \frac{2}{8}, \quad \frac{7}{8} = \frac{7}{8}$$ Sum and subtract: $$\frac{20}{8} + \frac{2}{8} - \frac{7}{8} = \frac{20 + 2 - 7}{8} = \frac{15}{8}$$ 3. **Divide by $-\frac{3}{16}$:** $$\frac{15}{8} : \left(-\frac{3}{16}\right) = \frac{15}{8} \times \left(-\frac{16}{3}\right)$$ Simplify numerator and denominator: $$= \frac{15}{\cancel{8}} \times \left(-\frac{16}{3}\right) = \frac{15}{\cancel{8}} \times \left(-\frac{16}{3}\right)$$ Cancel common factors: $$= \frac{15}{\cancel{8}} \times \left(-\frac{16}{3}\right) = \frac{15}{\cancel{8}} \times \left(-\frac{16}{3}\right)$$ Actually, $8$ and $16$ share factor $8$: $$= \frac{15}{8} \times \left(-\frac{16}{3}\right) = 15 \times \left(-\frac{2}{3}\right) = -\frac{30}{3} = -10$$ 4. **Calculate $-\frac{5}{3} \times \left(-\frac{2}{7}\right)$:** $$-\frac{5}{3} \times \left(-\frac{2}{7}\right) = \frac{10}{21}$$ 5. **Calculate $\left(-\frac{1}{2}\right) \times \left(-2\right)^3$:** First, compute $\left(-2\right)^3 = -8$. Then: $$\left(-\frac{1}{2}\right) \times (-8) = 4$$ 6. **Simplify $\left(3 - \frac{9}{5}\right) : \left(-\frac{1}{5}\right)$:** Convert 3 to fraction with denominator 5: $$3 = \frac{15}{5}$$ Subtract: $$\frac{15}{5} - \frac{9}{5} = \frac{6}{5}$$ Divide by $-\frac{1}{5}$: $$\frac{6}{5} : \left(-\frac{1}{5}\right) = \frac{6}{5} \times \left(-5\right) = -6$$ 7. **Combine all results:** $$-10 + \frac{10}{21} + 4 - (-6) = -10 + \frac{10}{21} + 4 + 6$$ Sum integers: $$-10 + 4 + 6 = 0$$ So expression reduces to: $$0 + \frac{10}{21} = \frac{10}{21}$$ **Final answer:** $$\boxed{\frac{10}{21}}$$