The solution set of the inequation

for $|x+2|>5$

We get,

$x+2>5$ or $x+2<-5$

i.e. $\quad x>3 \quad($ By adding $-2$ to both sides of $x+2>5)$

and $x<-7 \quad($ By adding $-2$ to both sides of $x+2<-5)$

Hence, solution set of $|x+2|>5$ is

$(-\infty,-7) \cup(3, \infty)$