$a)\underset{x\to 2^{+}}{\lim }\left(\frac{1}{{\left(x-2\right)}^2}-\frac{1}{x-4}\right)=\frac{1}{0^{+}}-\frac{1}{-2}=\frac{1}{0^{+}}+\frac{1}{2}=\infty +\frac{1}{2}=\infty$  

 

$b)\underset{x\to 2^{+}}{\lim }\left(\frac{1}{{\left(x-2\right)}^2}-\frac{1}{x^2-4}\right)\stackrel{\left[\infty -\infty \right]}{=}\underset{x\to 2^{+}}{\lim }\left(\frac{1}{{\left(x-2\right)}^2}-\frac{1}{\left(x-2\right)\left(x+2\right)}\right)=\underset{x\to 2^{+}}{\lim }\left(\frac{x+2}{{\left(x-2\right)}^2\left(x+2\right)}-\frac{x-2}{{\left(x-2\right)}^2\left(x+2\right)}\right)=\underset{x\to 2^{+}}{\lim }\frac{4}{{\left(x-2\right)}^2\left(x+2\right)}=\frac{4}{0^{+}\cdot 4}=\infty$