For what value(s) of a and b, $f\left(x\right)= \begin{cases}
\dfrac{x+2}{\left|x+2\right|}+a, & x<-2 \\
& \\
a+b, & x=-2 \\
& \\
\dfrac{x+2}{\left|x+2\right|}+b, & x>-2
\end{cases}$
is continuous at $x=-2$
$f\left(-2\right)=a+b$ $\cdots$ (1)
When $x<-2, \; \left|x+2\right|=-\left(x+2\right)$
When $x>-2, \; \left|x+2\right|=\left(x+2\right)$
$\begin{aligned}
\therefore \lim\limits_{x \to -2^-} f\left(x\right) & =\lim\limits_{x\to -2^-} \; \dfrac{x+2}{\left|x+2\right|}+a \\
& \\
& = \lim\limits_{x \to -2} \; \dfrac{x+2}{-\left(x+2\right)}+a \\
& \\
& = -1+a \;\;\; \cdots (2)
\end{aligned}$
$\begin{aligned}
\lim\limits_{x \to -2^+} f\left(x\right) & =\lim\limits_{x\to -2^+} \; \dfrac{x+2}{\left|x+2\right|}+b \\
& \\
& = \lim\limits_{x \to -2} \; \dfrac{x+2}{\left(x+2\right)}+b \\
& \\
& = 1+b \;\;\; \cdots (3)
\end{aligned}$
Since the function is continuous at $x=-2$,
$\lim\limits_{x \to -2^-}f\left(x\right)=f\left(-2\right)=\lim\limits_{x \to -2^+}f\left(x\right)$
i.e. $-1+a=a+b$ $\implies$ $b=-1$ [from equations (1) and (2)]
and $1+b=a+b$ $\implies$ $a=1$ [from equations (1) and (3)]