Ответ:

1) \[\frac{2y-7}{y^2-9}-\frac{y-10}{y^2-9}=\frac{2y-7-y+10}{y^2-9}=\frac{y+3}{(y-3)(y+3)}=\frac{1}{y-3}.\]

При \(y=3{,}1\): \[\frac{1}{3{,}1-3}=\frac{1}{0{,}1}=10.\]

При \(y=-2\): \[\frac{1}{-2-3}=-\frac{1}{5}.\]

2) Так как \(4-c^2=-(c^2-4)\), получаем:

\[-\frac{3c-5}{4-c^2}+\frac{3-2c}{c^2-4}=\frac{3c-5}{c^2-4}+\frac{3-2c}{c^2-4}=\frac{c-2}{(c-2)(c+2)}=\frac{1}{c+2}.\]

При \(c=3\): \[\frac{1}{3+2}=\frac{1}{5}.\]

При \(c=-3\): \[\frac{1}{-3+2}=-1.\]

Ответ: 1) \(10; -\frac{1}{5}\); 2) \(\frac{1}{5}; -1\).