1. Stating the problem: Simplify the complex numbers $Z_1 = \frac{-1}{i}$ and $Z_2 = \frac{1}{2} + i\frac{\sqrt{3}}{2} \div -\sqrt{3} - i$.\n\n2. Simplify $Z_1$: Recall that $\frac{1}{i} = -i$ since $i \times (-i) = 1$. Therefore, $Z_1 = \frac{-1}{i} = -1 \times (-i) = i$.\n\n3. Simplify $Z_2$: Write explicitly as $Z_2 = \frac{1}{2} + \frac{i\sqrt{3}}{2} \div (-\sqrt{3} - i)$.\nFirst, rewrite division as multiplication by reciprocal: $Z_2 = \frac{1}{2} + \frac{i\sqrt{3}}{2} \times \frac{1}{-\sqrt{3} - i}$.\n\n4. Rationalize the denominator for the reciprocal $\frac{1}{-\sqrt{3} - i}$ by multiplying numerator and denominator by conjugate $-\sqrt{3} + i$: $$\frac{1}{-\sqrt{3} - i} \times \frac{-\sqrt{3} + i}{-\sqrt{3} + i} = \frac{-\sqrt{3} + i}{(-\sqrt{3})^2 - (i)^2} = \frac{-\sqrt{3} + i}{3 - (-1)} = \frac{-\sqrt{3} + i}{4}.$$\n\n5. Substitute back into $Z_2$: $$Z_2 = \frac{1}{2} + \frac{i\sqrt{3}}{2} \times \frac{-\sqrt{3} + i}{4} = \frac{1}{2} + \frac{i\sqrt{3}(-\sqrt{3} + i)}{8}.$$\n\n6. Simplify the numerator inside the product: $$i\sqrt{3}(-\sqrt{3} + i) = i\sqrt{3}(-\sqrt{3}) + i\sqrt{3}(i) = -i3 + i^2 \sqrt{3} = -3i - \sqrt{3}$$ because $i^2 = -1$.\n\n7. Plug back into expression: $$Z_2 = \frac{1}{2} + \frac{-3i - \sqrt{3}}{8} = \frac{1}{2} - \frac{\sqrt{3}}{8} - \frac{3i}{8}.$$\n\n8. Write final simplified form for $Z_2$: $$Z_2 = \left(\frac{1}{2} - \frac{\sqrt{3}}{8}\right) - \frac{3}{8}i.$$\n\n9. Final answers: $$ Z_1 = i, \quad Z_2 = \frac{1}{2} - \frac{\sqrt{3}}{8} - \frac{3}{8}i.$$