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

To simplify the first expression, start by rewriting everything in terms of prime factors. Notice that \( 15 = 3 \cdot 5 \) and \( 45 = 3^2 \cdot 5 \). This means: \[ 15^{n+2} = (3 \cdot 5)^{n+2} = 3^{n+2} \cdot 5^{n+2} \] \[ 45^{1-n} = (3^2 \cdot 5)^{1-n} = 3^{2(1-n)} \cdot 5^{1-n} \] \[ 3^{3-n} = 3^{3-n} \] Now substituting these back into the expression: \[ \frac{3^{n+2} \cdot 5^{n+2} \cdot 3^{2(1-n)} \cdot 5^{1-n}}{3^{3-n}} \] Combining the powers of \(3\) and \(5\): \[ = \frac{3^{n+2 + 2(1-n)} \cdot 5^{n+2 + 1-n}}{3^{3-n}} = \frac{3^{n+2 + 2 - 2n} \cdot 5^{n+3-n}}{3^{3-n}} = \frac{3^{4-n} \cdot 5^{n+3}}{3^{3-n}} \] This simplifies to: \[ 3^{(4-n)-(3-n)} \cdot 5^{n+3} = 3^{1} \cdot 5^{n+3} = 3 \cdot 5^{n+3} \] --- For the second expression, begin by factoring \(2^x\) from the numerator: \[ \frac{2^{x} + 2^{x + 1}}{2^{x} - 2^{x + 2}} = \frac{2^x (1 + 2)}{2^x (1 - 4)} \] Now, simplifying further: \[ = \frac{2^x \cdot 3}{2^x \cdot (-3)} = \frac{3}{-3} = -1 \] So the simplified expression is: \[ -1 \]