1.3.4. \( \left(7^{-2}+7^{-1}\right) \div(21 \div 3) \)

Let's simplify the expression step by step. First, we compute the innermost operation: \[ 21 \div 3 = 7. \] Next, we handle the exponent terms: \[ 7^{-2} = \frac{1}{7^2} = \frac{1}{49}, \quad 7^{-1} = \frac{1}{7}. \] Now, we add these two fractions: \[ 7^{-2} + 7^{-1} = \frac{1}{49} + \frac{1}{7}. \] To add these, we need a common denominator. The least common denominator is 49. We can convert \(\frac{1}{7}\) as follows: \[ \frac{1}{7} = \frac{7}{49}. \] Now we can perform the addition: \[ \frac{1}{49} + \frac{7}{49} = \frac{8}{49}. \] Now we can complete the original expression: \[ (7^{-2} + 7^{-1}) \div (21 \div 3) = \frac{8}{49} \div 7. \] Dividing by 7 is the same as multiplying by \(\frac{1}{7}\): \[ \frac{8}{49} \div 7 = \frac{8}{49} \cdot \frac{1}{7} = \frac{8}{343}. \] So the final answer is \[ \frac{8}{343}. \]