1. **State the problem:** Find the exact value of the magnitude $$\left|\frac{4-2i}{-i^5(2-3i)}\right|$$ without using a calculator. 2. **Recall the formula for magnitude of a quotient:** For any complex numbers $z$ and $w$, $$\left|\frac{z}{w}\right| = \frac{|z|}{|w|}$$ 3. **Calculate the magnitude of the numerator:** $$|4-2i| = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$ 4. **Simplify the denominator:** First, simplify $-i^5$. Recall that $i^4 = 1$, so $$i^5 = i^{4} \cdot i = 1 \cdot i = i$$ Therefore, $$-i^5 = -i$$ 5. **Calculate the magnitude of the denominator:** $$|-i(2-3i)| = | -i| \cdot |2-3i| = |i| \cdot \sqrt{2^2 + (-3)^2} = 1 \cdot \sqrt{4 + 9} = \sqrt{13}$$ 6. **Combine the magnitudes:** $$\left|\frac{4-2i}{-i^5(2-3i)}\right| = \frac{|4-2i|}{|-i(2-3i)|} = \frac{2\sqrt{5}}{\sqrt{13}}$$ 7. **Rationalize the denominator:** $$\frac{2\sqrt{5}}{\sqrt{13}} = \frac{2\sqrt{5}}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}} = \frac{2\sqrt{65}}{13}$$ **Final answer:** $$\boxed{\frac{2\sqrt{65}}{13}}$$