Answer
1) The open interval is $(-10/9,-10/11)$
2) $\delta=1/11$
Work Step by Step
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-(-1)|\lt\delta\Rightarrow|\frac{1}{x}-(-1)|\lt0.1$$
1) Find the interval around $-1$ on which $|\frac{1}{x}-(-1)|\lt0.1$ holds.
Solve the inequality: $$|\frac{1}{x}-(-1)|\lt0.1$$ $$|\frac{1}{x}+1|\lt0.1$$ $$-0.1\lt\frac{1}{x}+1\lt0.1$$ $$-1.1\lt\frac{1}{x}\lt-0.9$$ $$-\frac{1}{1.1}\gt x\gt-\frac{1}{0.9}$$ $$-\frac{10}{11}\gt x\gt-\frac{10}{9}$$
The open interval around $-1$ is $(-10/9,-10/11)$.
2) Give a value for $\delta$
The nearer endpoint to $-1$ is $-10/11$, and the distance between them is $|-1|-|-10/11|=1/11$.
So if we take $\delta=1/11$ or any smaller positive number, then $0\lt|x-(-1)|\lt1/11$, meaning all $x$ would be placed in the interval $(-10/11,-10/9)$ so that $|\frac{1}{x}-(-1)|\lt0.1$.
In other words, $$0\lt |x-(-1)|\lt1/11\Rightarrow|\frac{1}{x}-(-1)|\lt0.1$$
