1. **Problem a:** Simplify $$\left(\frac{4^{3}}{4^{-4}}\right)^{2 \cdot \frac{1}{2}}$$. 2. Use the property of exponents for division: $$\frac{a^{m}}{a^{n}} = a^{m-n}$$. 3. Simplify inside the parentheses: $$\frac{4^{3}}{4^{-4}} = 4^{3 - (-4)} = 4^{3 + 4} = 4^{7}$$. 4. The exponent outside is $$2 \cdot \frac{1}{2} = 1$$, so: $$\left(4^{7}\right)^{1} = 4^{7}$$. 5. **Final answer for a:** $$4^{7}$$. 1. **Problem b:** Simplify $$\frac{\sqrt{4^{3}}}{4^{-5}} \cdot \sqrt[6]{4^{2}} \cdot (4^{2})^{4} \cdot 4^{3}$$. 2. Rewrite radicals as fractional exponents: $$\sqrt{4^{3}} = (4^{3})^{\frac{1}{2}} = 4^{\frac{3}{2}}$$, $$\sqrt[6]{4^{2}} = (4^{2})^{\frac{1}{6}} = 4^{\frac{2}{6}} = 4^{\frac{1}{3}}$$. 3. Simplify powers: $$(4^{2})^{4} = 4^{2 \cdot 4} = 4^{8}$$. 4. Substitute back: $$\frac{4^{\frac{3}{2}}}{4^{-5}} \cdot 4^{\frac{1}{3}} \cdot 4^{8} \cdot 4^{3}$$. 5. Use division property: $$\frac{4^{\frac{3}{2}}}{4^{-5}} = 4^{\frac{3}{2} - (-5)} = 4^{\frac{3}{2} + 5} = 4^{\frac{3}{2} + \frac{10}{2}} = 4^{\frac{13}{2}}$$. 6. Now multiply all powers of 4: $$4^{\frac{13}{2}} \cdot 4^{\frac{1}{3}} \cdot 4^{8} \cdot 4^{3} = 4^{\frac{13}{2} + \frac{1}{3} + 8 + 3}$$. 7. Convert all to common denominator 6: $$\frac{13}{2} = \frac{39}{6}, \quad \frac{1}{3} = \frac{2}{6}, \quad 8 = \frac{48}{6}, \quad 3 = \frac{18}{6}$$. 8. Sum exponents: $$\frac{39}{6} + \frac{2}{6} + \frac{48}{6} + \frac{18}{6} = \frac{39 + 2 + 48 + 18}{6} = \frac{107}{6}$$. 9. **Final answer for b:** $$4^{\frac{107}{6}}$$. 1. **Problem c:** Simplify $$\frac{4^{-8}}{4^{-10}} \div \left( \sqrt[5]{4^{6}} \right)^{0}$$. 2. Use division property: $$\frac{4^{-8}}{4^{-10}} = 4^{-8 - (-10)} = 4^{-8 + 10} = 4^{2}$$. 3. Any number to the zero power is 1: $$\left( \sqrt[5]{4^{6}} \right)^{0} = 1$$. 4. So the expression becomes: $$4^{2} \div 1 = 4^{2}$$. 5. **Final answer for c:** $$4^{2}$$.