11. (35分)计算:
(1)$\frac {7}{5}×(\frac {1}{3}-\frac {1}{2})×\frac {3}{7}÷\frac {5}{4}$
(2)$-18÷(+3.25)÷(-2\frac {1}{4})$
(3)$[82\frac {2}{3}-6÷(-3^{2})]×(-\frac {2}{5})^{3}÷(-8)$

（1）$-\frac{2}{25}$ （2）$\frac{32}{13}$ （3）$\frac{2}{3}$

(1) $\frac{7}{5} × \left(\frac{1}{3} - \frac{1}{2}\right) × \frac{3}{7} ÷ \frac{5}{4}$
$=\frac{7}{5} × \left(-\frac{1}{6}\right) × \frac{3}{7} × \frac{4}{5}$
$=\frac{7}{5} × \frac{3}{7} × \left(-\frac{1}{6}\right) × \frac{4}{5}$
$=\frac{3}{5} × \left(-\frac{2}{15}\right)$
$=-\frac{6}{75}$
$=-\frac{2}{25}$
(2) $-18 ÷ (+3.25) ÷ \left(-2\frac{1}{4}\right)$
$=-18 ÷ \frac{13}{4} ÷ \left(-\frac{9}{4}\right)$
$=-18 × \frac{4}{13} × \left(-\frac{4}{9}\right)$
$=-18 × \left(-\frac{4}{9}\right) × \frac{4}{13}$
$=8 × \frac{4}{13}$
$=\frac{32}{13}$
(3) $\left[82\frac{2}{3} - 6 ÷ (-3^2)\right] × \left(-\frac{2}{5}\right)^3 ÷ (-8)$
$=\left[\frac{248}{3} - 6 ÷ (-9)\right] × \left(-\frac{8}{125}\right) ÷ (-8)$
$=\left[\frac{248}{3} + \frac{2}{3}\right] × \left(-\frac{8}{125}\right) × \left(-\frac{1}{8}\right)$
$=\frac{250}{3} × \left[\left(-\frac{8}{125}\right) × \left(-\frac{1}{8}\right)\right]$
$=\frac{250}{3} × \frac{1}{125}$
$=\frac{2}{3}$

12. (14分)观察下列各式:
$\frac {1}{2}×\frac {2}{3}= \frac {1}{3},\frac {1}{2}×\frac {2}{3}×\frac {3}{4}= \frac {1}{4},\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×\frac {4}{5}= \frac {1}{5},...$
(1)猜想:$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×... ×\frac {49}{50}= $______;
(2)根据上面的规律,计算:$(\frac {1}{100}-1)×(\frac {1}{99}-1)×(\frac {1}{98}-1)×... ×(\frac {1}{2}-1)$.

（1）$\frac{1}{50}$ （2）原式$=-\frac{99}{100}×(-\frac{98}{99})×(-\frac{97}{98})×\cdots×(-\frac{1}{2})=-\frac{1}{100}$

(1) $\frac{1}{50}$
(2) 原式$=(-\frac{99}{100})×(-\frac{98}{99})×(-\frac{97}{98})×\cdots×(-\frac{1}{2})$
$=-\frac{99}{100}×\frac{98}{99}×\frac{97}{98}×\cdots×\frac{1}{2}$
$=-\frac{1}{100}$

13. (11分)我们知道$a÷b= \frac {a}{b},b÷a= \frac {b}{a}$,显然$a÷b与b÷a$的结果是互为倒数的关系.小明利用这一思想方法计算$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})$的过程如下:因为原式的倒数为$(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})÷(-\frac {1}{30})= (\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})×(-30)= -20+3-5+12= -10$,所以$(-\frac {1}{30})÷(\frac {2}{3}-\frac {1}{10}+\frac {1}{6}-\frac {2}{5})= -\frac {1}{10}$.
请你仿照这种方法计算:$(-\frac {1}{42})÷(\frac {1}{6}-\frac {3}{14}+\frac {2}{3}-\frac {2}{7})$.

因为原式的倒数为$(\frac{1}{6}-\frac{3}{14}+\frac{2}{3}-\frac{2}{7})÷(-\frac{1}{42})=(\frac{1}{6}-\frac{3}{14}+\frac{2}{3}-\frac{2}{7})×(-42)=-7+9-28+12=-14$，所以$(-\frac{1}{42})÷(\frac{1}{6}-\frac{3}{14}+\frac{2}{3}-\frac{2}{7})=-\frac{1}{14}$