1. **Evaluate** $\log_9 27$: Recall $\log_a b = c$ means $a^c = b$. We rewrite bases: $9 = 3^2$, $27 = 3^3$. So $\log_9 27 = x$ means $(3^2)^x = 3^3$. This simplifies to $3^{2x} = 3^3$, so $2x = 3$ and $x = \frac{3}{2}$. 2. **Simplify** $\frac{5^{18}}{5^{13}}$: Use the rule $\frac{a^m}{a^n} = a^{m-n}$. So $5^{18 - 13} = 5^5$. 3. **Calculate** $\log_5 0.25$: Recall $0.25 = \frac{1}{4} = 4^{-1}$ and $4 = 2^2$. Express 0.25 as power of 5: $0.25 = \frac{1}{4}$, harder with 5 directly, use common logs or rewrite 0.25 as $5^{-y}$: Because $\log_5 0.25 = y$, means $5^y = 0.25$. We know $0.25 = \frac{1}{4}$, so rewrite $4$ as $5^{\log_5 4}$. Alternatively, note $0.25 = \frac{1}{4} = \frac{1}{2^2}$. Since 5 and 2 have no simple relation, use decimal approximation: $\log_5 0.25 = \frac{\log 0.25}{\log 5} \approx \frac{-0.60206}{0.69897} \approx -0.861$. 4. **Simplify** $2x^{-2} \times 5x$: Rewrite powers: $2x^{-2} \times 5x = 2 \times 5 \times x^{-2} \times x^1 = 10 x^{(-2 + 1)} = 10 x^{-1} = \frac{10}{x}$. 5. **Simplify** $0.8^{\frac{2}{3}}$: Rewrite 0.8 as fraction: $0.8 = \frac{4}{5}$. So, $\left(\frac{4}{5}\right)^{\frac{2}{3}} = \left(\left(\frac{4}{5}\right)^{\frac{1}{3}}\right)^2$. That means cube root of $\frac{4}{5}$, squared. 6. **Simplify** $\sqrt{1 - \frac{9}{16}}$: Calculate inside the root: $1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16}$. So $\sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$. **Final answers:** 1. $\log_9 27 = \frac{3}{2}$ 2. $\frac{5^{18}}{5^{13}} = 5^5$ 3. $\log_5 0.25 \approx -0.861$ 4. $2x^{-2} \times 5x = \frac{10}{x}$ 5. $0.8^{\frac{2}{3}} = \left(\left(\frac{4}{5}\right)^{\frac{1}{3}}\right)^2$ 6. $\sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$