$a)\underset{x\to 2}{\lim }\frac{x^3+x^2+3}{x^2-1}=\frac{2^3+2^2+3}{2^2-1}=\frac{8+4+3}{3}=\frac{15}{3}=5$  

 

$b)\underset{x\to -1}{\lim }\frac{x^4-3x+8}{4x-x^3}=\frac{{\left(-1\right)}^4-3\cdot \left(-1\right)+8}{4\cdot \left(-1\right)-{\left(-1\right)}^3}=\frac{1+3+8}{-4+1}=\frac{12}{-3}=-4$  

 

$c)\underset{x\to -3}{\lim }\frac{{\left(x+2\right)}^5}{7-x^2}=\frac{{\left(-3+2\right)}^5}{\left(7-{\left(-3\right)}^2\right)}=\frac{-1}{7-9}=\frac{-1}{-2}=\frac{1}{2}$