Verify that infinite series is equal the given result with summation technique

$($a$){\displaystyle \underset{n=-}{\overset{}{}}}\frac{1}{(2n-3{)}^2}=\frac{{}^2}{4}$

$($b$){\displaystyle \underset{n=-}{\overset{}{}}}\frac{1}{n^2+9}=\frac{}{3}coth(3)$

$($c$){\displaystyle \underset{n=-}{\overset{}{}}}\frac{(-1{)}^n}{(n+a{)}^2}=\frac{{}^{2}cos(a)}{(sin(a){)}^2}$

$($d$){\displaystyle \underset{n=-}{\overset{}{}}}\frac{1}{(n-\frac{1}{2}{)}^2}=\frac{{}^2}{2}$

$($e$){\displaystyle \underset{n=-}{\overset{}{}}}\frac{(-1{)}^n}{n^2}=\frac{-{}^2}{12}$