1. **Statement of the problem:** Calculate the following expressions: $$\frac{1}{3^{-2}}; \quad (-1)^{-49}; \quad 0.18$$ 2. **Formulas and rules:** - For any nonzero number $a$ and integer $n$, $a^{-n} = \frac{1}{a^n}$. - Powers of $-1$: $(-1)^n = -1$ if $n$ is odd, and $1$ if $n$ is even. 3. **Step-by-step calculations:** **First expression:** $$\frac{1}{3^{-2}} = 3^{2} = 9$$ Explanation: Since $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$, its reciprocal is $9$. **Second expression:** $$(-1)^{-49} = \frac{1}{(-1)^{49}}$$ Since $49$ is odd, $(-1)^{49} = -1$, so: $$(-1)^{-49} = \frac{1}{-1} = -1$$ **Third expression:** $0.18$ is a decimal number and remains as is. 4. **Final answers:** $$\frac{1}{3^{-2}} = 9; \quad (-1)^{-49} = -1; \quad 0.18 = 0.18$$