\( \begin{array}{ll}\text { 1.1.2. } & \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} \\ \text { 1.1.3 } & \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}}\end{array} \)

\[ \textbf{1.1.2: } \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} \] Write \(15\) and \(45\) in terms of their prime factors: \[ 15 = 3 \cdot 5,\quad 45 = 3^2 \cdot 5. \] Rewrite the numerator: \[ 15^{n+2} = (3\cdot5)^{n+2} = 3^{n+2} \cdot 5^{n+2}, \] \[ 45^{1-n} = (3^2\cdot5)^{1-n} = 3^{2(1-n)} \cdot 5^{1-n}. \] Multiply these: \[ 15^{n+2} \cdot 45^{1-n} = 3^{n+2} \cdot 5^{n+2} \cdot 3^{2(1-n)} \cdot 5^{1-n}. \] Combine the powers of \(3\) and \(5\): \[ 3^{n+2} \cdot 3^{2(1-n)} = 3^{n+2 + 2(1-n)} = 3^{n+2 + 2 - 2n} = 3^{4-n}, \] \[ 5^{n+2} \cdot 5^{1-n} = 5^{n+2+1-n} = 5^{3}. \] Thus, the numerator becomes: \[ 3^{4-n} \cdot 5^{3}. \] Now, divide by the denominator: \[ \frac{3^{4-n} \cdot 5^{3}}{3^{3-n}}. \] Simplify the powers of \(3\): \[ \frac{3^{4-n}}{3^{3-n}} = 3^{(4-n) - (3-n)} = 3^{1} = 3. \] So the entire expression becomes: \[ 3 \cdot 5^3 = 3 \cdot 125 = 375. \] \[ \textbf{1.1.3: } \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}} \] Factor \(2^x\) from each term. For the numerator: \[ 2^{x}+2^{x+1} = 2^x (1+2) = 2^x \cdot 3. \] For the denominator: \[ 2^{x}-2^{x+2} = 2^x (1-2^2) = 2^x (1-4) = 2^x \cdot (-3). \] Now, the fraction becomes: \[ \frac{2^x \cdot 3}{2^x \cdot (-3)}. \] Cancel the common factor \(2^x\) (assuming \(2^x \neq 0\)): \[ \frac{3}{-3} = -1. \] Thus, the simplified results are: 1. \( \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} = 375 \) 2. \( \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}} = -1 \)