1.2. Between which two consecutive integers does \( \sqrt{271} \) lie? 1.3. Simplify the following without the aid of a calculator: 1.3.1. \( \left(3,6 \times 10^{6}\right)-\left(5,2 \times 10^{5}\right) \) 1.3.2. \( \frac{1}{2}+2 \frac{3}{4} \cdot \frac{3}{8} \)

To find the consecutive integers between which \( \sqrt{271} \) lies, we can start by knowing that \( 16^2 = 256 \) and \( 17^2 = 289 \). Since \( 256 < 271 < 289 \), we can conclude that \( \sqrt{271} \) lies between 16 and 17. When simplifying \( (3.6 \times 10^6) - (5.2 \times 10^5) \), it's helpful to express the second term with the same power of ten. We can rewrite \( 5.2 \times 10^5 \) as \( 0.52 \times 10^6 \). Now, we have \( (3.6 - 0.52) \times 10^6 = 3.08 \times 10^6 \). For the expression \( \frac{1}{2} + 2 \frac{3}{4} \cdot \frac{3}{8} \), first, convert \( 2 \frac{3}{4} \) into an improper fraction: \( \frac{11}{4} \). Now, multiplying \( \frac{11}{4} \cdot \frac{3}{8} = \frac{33}{32} \). Adding \( \frac{1}{2} \) (which is \( \frac{16}{32} \)) gives us \( \frac{16}{32} + \frac{33}{32} = \frac{49}{32} \). So the final answer is \( \frac{49}{32} \), or \( 1 \frac{17}{32} \).