1. The domain of the function $f(x)$ is $D$. If
(1) $f(x)$ is a monotonic function in $D$,
(2) There exists an interval $[a, b] \subseteq D$ such that the range of $f(x)$ on the interval $[a, b]$ is $[a, b]$,
then $y=f(x)$ is called a closed function.
If $f(x)=\sqrt{x+2}+k$ is a closed function, then the range of values for $k$ is
$-1 .\left(-\frac{9}{4},-2\right]$.
Notice that, $\sqrt{x+2}+k=x$ has two distinct real roots in the interval $[-2,+\infty)$.
Let $\sqrt{x+2}=t \in[0,+\infty)$.
Then $g(t)=t^{2}-t-(k+2)=0$ has two distinct real roots in the interval $[0,+\infty)$.
Thus, $\Delta=1+4(k+2)>0$, and $g(0) \geqslant 0$.
Solving this, we get $k \in\left(-\frac{9}{4},-2\right]$.