Answer
a. $\infty$
b. Does not exist.
Work Step by Step
a. With $x\to0,x\ne0, x\ne-1, x+1\gt0$, we have $\lim_{x\to0}\frac{x^2+x}{x^5+2x^4+x^3}=\lim_{x\to0}\frac{1}{x^2}\frac{x+1}{x^2+2x+1}=\lim_{x\to0}\frac{1}{x^2}\frac{1}{x+1}=\infty$
b. With $x\to-1,x\ne0, x\ne-1$, we have $\lim_{x\to-1}\frac{x^2+x}{x^5+2x^4+x^3}=\lim_{x\to-1}\frac{1}{x^2}\frac{x+1}{x^2+2x+1}=\lim_{x\to-1}\frac{1}{x^2}\frac{1}{x+1}$.
Since $\lim_{x\to-1^+}\frac{1}{x^2}\frac{1}{x+1}=\infty$ and $\lim_{x\to-1^-}\frac{1}{x^2}\frac{1}{x+1}=-\infty$, the limit does not exist at $x\to-1$
