7. 计算:$(-36\frac{9}{10})÷ 9= $.

$-4\frac{1}{10}$

$(-36\frac{9}{10})÷9 = (-36 - \frac{9}{10})×\frac{1}{9} = -36×\frac{1}{9} - \frac{9}{10}×\frac{1}{9} = -4 - \frac{1}{10} = -4\frac{1}{10}$

8. 定义$a★b= -\frac{1}{a}÷ \frac{b}{2}$,例如,$2★3= -\frac{1}{2}÷ \frac{3}{2}= -\frac{1}{3}$,则$\frac{2}{3}★\frac{9}{4}$的值为.

$-\frac{4}{3}$

根据题目中给出的运算规则$a★b= -\frac{1}{a} ÷ \frac{b}{2}$，
我们可以将$\frac{2}{3}★\frac{9}{4}$转化为标准的数学运算。
即：
$\frac{2}{3}★\frac{9}{4} = -\frac{1}{\frac{2}{3}} ÷ \frac{\frac{9}{4}}{2}$
$= -\frac{3}{2} ÷ \frac{9}{8}$
$= -\frac{3}{2} × \frac{8}{9}$
$= -\frac{4}{3}$

9. 计算.
(1)$-2.5÷ \frac{5}{8}× (-\frac{1}{4})$;
(2)$-27÷ 2\frac{1}{4}× \frac{4}{9}÷ (-24)$;
(3)$(-\frac{3}{5})× (-3\frac{1}{2})÷ (-1\frac{1}{4})÷ 3$;
(4)$-5÷ (-1\frac{2}{7})× \frac{4}{5}× (-2\frac{1}{4})÷ 7$.

(1)
$-2.5 ÷ \frac{5}{8} × (-\frac{1}{4})$
$= -2.5 × \frac{8}{5} × (-\frac{1}{4})$
$= -4 × (-\frac{1}{4})$
$= 1$
(2)
$-27 ÷ 2\frac{1}{4} × \frac{4}{9} ÷ (-24)$
$= -27 ÷ \frac{9}{4} × \frac{4}{9} × (-\frac{1}{24})$
$= -27 × \frac{4}{9} × \frac{4}{9} × (-\frac{1}{24})$
$= -12 × \frac{4}{9} × (-\frac{1}{24})$
$= -\frac{16}{3} × (-\frac{1}{24})$
$= \frac{2}{9}$
(3)
$(-\frac{3}{5}) × (-3\frac{1}{2}) ÷ (-1\frac{1}{4}) ÷ 3$
$= (-\frac{3}{5}) × (-\frac{7}{2}) × (-\frac{4}{5}) × \frac{1}{3}$
$= \frac{21}{10} × (-\frac{4}{5}) × \frac{1}{3}$
$= -\frac{42}{25} × \frac{1}{3}$
$= -\frac{14}{25}$
(4)
$-5 ÷ (-1\frac{2}{7}) × \frac{4}{5} × (-2\frac{1}{4}) ÷ 7$
$= -5 ÷ (-\frac{9}{7}) × \frac{4}{5} × (-\frac{9}{4}) × \frac{1}{7}$
$= -5 × (-\frac{7}{9}) × \frac{4}{5} × (-\frac{9}{4}) × \frac{1}{7}$
$= \frac{35}{9} × \frac{4}{5} × (-\frac{9}{4}) × \frac{1}{7}$
$= \frac{28}{9}× (-\frac{9}{4}) × \frac{1}{7}$
$= -7 × \frac{1}{7}$
$= -1$

10. 计算$6÷ (-\frac{1}{2}+\frac{1}{3})$时,方方同学的计算过程如下:原式$=6÷ (-\frac{1}{2})+6÷ \frac{1}{3}= -12+18= 6$.请判断方方的计算过程是否正确.若不正确,请写出正确的计算过程.

不正确。
正确的计算过程为：
首先计算括号内的有理数加法：
$-\frac{1}{2} + \frac{1}{3} = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6}$
然后，进行除法运算：
$6 ÷ (-\frac{1}{6}) = 6 × (-6) = -36$
所以，$6÷ (-\frac{1}{2}+\frac{1}{3}) = -36$。