$$
\begin{array}{l}
\text { 1. Let } f(x)=x^{2}+a x+b \cos x \text {, and } \\
\{x \mid f(x)=0, x \in \mathbf{R}\} \\
=\{x \mid f(f(x))=0, x \in \mathbf{R}\} \neq \varnothing .
\end{array}
$$

Then the range of values for $a+b$ is $\qquad$
-、1. $[0,4)$.
Let $x_{0} \in\{x \mid f(x)=0, x \in \mathbf{R}\}$.
Then $b=f(0)=f\left(f\left(x_{0}\right)\right)=0$.
Thus, $f(x)=x(x+a)$.
Therefore, $f(f(x))=x(x+a)\left(x^{2}+a x+a\right)$.
When $a=0$, it clearly satisfies the condition;
When $a \neq 0$, since $0$ and $-a$ are not roots of $x^{2}+a x+a=0$, we have $a^{2}-4 a<0$, which means $0<a<4$.
In conclusion, $a+b \in[0,4)$.