Задание состоит из двух примеров.
\[ 3\frac{5}{7} = \frac{3 \cdot 7 + 5}{7} = \frac{21+5}{7} = \frac{26}{7} \]
\[ 1\frac{1}{7} = \frac{1 \cdot 7 + 1}{7} = \frac{7+1}{7} = \frac{8}{7} \]
\[ 4\frac{2}{7} = \frac{4 \cdot 7 + 2}{7} = \frac{28+2}{7} = \frac{30}{7} \]
\[ \frac{26}{7} - \frac{8}{7} + \frac{30}{7} = \frac{26 - 8 + 30}{7} = \frac{18 + 30}{7} = \frac{48}{7} \]
\[ \frac{48}{7} = 6\frac{6}{7} \]
\[ 1 - \frac{2}{7} = \frac{7}{7} - \frac{2}{7} = \frac{7-2}{7} = \frac{5}{7} \]
\[ 3\frac{5}{7} - \frac{5}{7} \]
\[ 3\frac{5}{7} = \frac{3 \cdot 7 + 5}{7} = \frac{21+5}{7} = \frac{26}{7} \]
\[ \frac{26}{7} - \frac{5}{7} = \frac{26 - 5}{7} = \frac{21}{7} \]
\[ \frac{21}{7} = 3 \]
Ответ: 1. \( 6\frac{6}{7} \); 2. \( 3 \).