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. \( \frac{1}{2}+2 \frac{3}{4} \cdot \frac{3}{8} \)

1.2. We first find two consecutive integers \(a\) and \(b\) such that \[ a^2 < 271 < b^2. \] Try \(a = 16\) and \(b = 17\): \[ 16^2 = 256 \quad \text{and} \quad 17^2 = 289. \] Since \[ 256 < 271 < 289, \] we conclude that \[ \sqrt{271} \text{ lies between } 16 \text{ and } 17. \] 1.3.1. We need to simplify \[ \left(3.6 \times 10^6\right) - \left(5.2 \times 10^5\right). \] Express \(5.2 \times 10^5\) with the same exponent as \(3.6 \times 10^6\). Since \[ 5.2 \times 10^5 = 0.52 \times 10^6, \] the subtraction becomes \[ 3.6 \times 10^6 - 0.52 \times 10^6 = (3.6 - 0.52) \times 10^6. \] Calculating the difference: \[ 3.6 - 0.52 = 3.08. \] Thus, the simplified form is \[ 3.08 \times 10^6. \] 1.3.2. We simplify the expression \[ \frac{1}{2} + 2\frac{3}{4} \cdot \frac{3}{8}. \] First, convert the mixed number \(2\frac{3}{4}\) into an improper fraction: \[ 2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8+3}{4} = \frac{11}{4}. \] Now, multiply: \[ \frac{11}{4} \cdot \frac{3}{8} = \frac{33}{32}. \] Then, add \(\frac{1}{2}\) to the product. Write \(\frac{1}{2}\) with a denominator of 32: \[ \frac{1}{2} = \frac{16}{32}. \] So, the sum is \[ \frac{16}{32} + \frac{33}{32} = \frac{49}{32}. \] Thus, the answers are: - \( \sqrt{271} \) lies between 16 and 17. - \( \left(3.6 \times 10^6\right) - \left(5.2 \times 10^5\right) = 3.08 \times 10^6 \). - \( \frac{1}{2}+2\frac{3}{4} \cdot \frac{3}{8} = \frac{49}{32} \).