Для решения примера раскроем скобки и приведём все дроби к общему знаменателю.
\[ 64\frac{2}{3} = \frac{64 \cdot 3 + 2}{3} = \frac{192 + 2}{3} = \frac{194}{3} \]
\[ 3\frac{1}{8} = \frac{3 \cdot 8 + 1}{8} = \frac{24 + 1}{8} = \frac{25}{8} \]
\[ 7\frac{1}{2} = \frac{7 \cdot 2 + 1}{2} = \frac{14 + 1}{2} = \frac{15}{2} \]
\[ 4\frac{1}{4} = \frac{4 \cdot 4 + 1}{4} = \frac{16 + 1}{4} = \frac{17}{4} \]
\[ 6\frac{1}{6} = \frac{6 \cdot 6 + 1}{6} = \frac{36 + 1}{6} = \frac{37}{6} \]
\[ 8\frac{1}{8} = \frac{8 \cdot 8 + 1}{8} = \frac{64 + 1}{8} = \frac{65}{8} \]
\[ \frac{194}{3} + \frac{25}{8} - (\frac{15}{2} + \frac{17}{4} - \frac{37}{6} - \frac{65}{8}) \]
\[ \frac{194}{3} + \frac{25}{8} - \frac{15}{2} - \frac{17}{4} + \frac{37}{6} + \frac{65}{8} \]
\[ \frac{194 \cdot 8}{3 \cdot 8} + \frac{25 \cdot 3}{8 \cdot 3} - \frac{15 \cdot 12}{2 \cdot 12} - \frac{17 \cdot 6}{4 \cdot 6} + \frac{37 \cdot 4}{6 \cdot 4} + \frac{65 \cdot 3}{8 \cdot 3} \]
\[ \frac{1552}{24} + \frac{75}{24} - \frac{180}{24} - \frac{102}{24} + \frac{148}{24} + \frac{195}{24} \]
\[ \frac{1552 + 75 - 180 - 102 + 148 + 195}{24} = \frac{1970 - 282}{24} = \frac{1688}{24} \]
\[ \frac{1688 \div 8}{24 \div 8} = \frac{211}{3} \]
\[ \frac{211}{3} = 70\frac{1}{3} \]
Ответ: $$70\frac{1}{3}$$