$$
\begin{array}{l}
\text { II. (40 points) Let } f(x)=x^{2}+a, \text { and define } \\
f^{1}(x)=f(x), \\
f^{n}(x)=f\left(f^{n-1}(x)\right)(n=2,3, \cdots) .
\end{array}
$$

Find the set
$$
M=\left\{a \in \mathbf{R}|| f^{n}(0) \mid \leqslant 2, n \in \mathbf{Z}_{+}\right\}
$$
Define the sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{1}=a, a_{n}=a_{n-1}^{2}+a(n \geqslant 2) \text {. }
$$

First, from $\left|a_{1}\right|=|a| \leqslant 2$, we get $-2 \leqslant a \leqslant 2$.
Thus, $M \subseteq[-2,2]$.
Second, by
$$
\begin{array}{l}
a_{k}-a_{k-1}=\left(a_{k-1}-\frac{1}{2}\right)^{2}+a-\frac{1}{4} \\
\geqslant a-\frac{1}{4}(k \geqslant 2),
\end{array}
$$

we have $a_{n}=\sum_{k=2}^{n}\left(a_{k}-a_{k-1}\right)+a_{1}$
$$
>(n-1)\left(a-\frac{1}{4}\right)+a \text {. }
$$

If $a>\frac{1}{4}$, then when $n>\frac{4(2-a)}{4 a-1}+1$, we have $a_{n}>2$.
Thus, $M \subseteq\left[-2, \frac{1}{4}\right]$.
(1) When $0 \leqslant a \leqslant \frac{1}{4}$,
$0 \leqslant a_{1}=a \leqslant \frac{1}{2}$,
$0 \leqslant a_{2}=a_{1}^{2}+a \leqslant\left(\frac{1}{2}\right)^{2}+\frac{1}{4}=\frac{1}{2}$.
Assume $0 \leqslant a_{k} \leqslant \frac{1}{2}$. Then
$$
0 \leqslant a_{k+1}=a_{k}^{2}+a \leqslant\left(\frac{1}{2}\right)^{2}+\frac{1}{4}=\frac{1}{2} \text {. }
$$

By mathematical induction, we know for any $n \in \mathbf{Z}_{+}$,
$$
0 \leqslant a_{n} \leqslant \frac{1}{2} \text {. }
$$
(2) When $-2 \leqslant a \leqslant 0$,
$$
\begin{array}{l}
a^{2}+2 a \leqslant 0,\left|a_{1}\right|=|a|, \\
-|a|=a \leqslant a_{2}=a_{1}^{2}+a \\
\leqslant a^{2}+a \leqslant-a=|a| .
\end{array}
$$

Thus, $\left|a_{2}\right| \leqslant|a|$.
Assume $\left|a_{k}\right| \leqslant|a|(k \geqslant 2)$. Then
$$
\begin{array}{l}
-|a|=a \leqslant a_{k+1}=a_{k}^{2}+a \\
\leqslant a^{2}+a \leqslant-a=|a| .
\end{array}
$$

Thus, $\left|a_{k+1}\right| \leqslant|a|$.
By mathematical induction, we know for any $n \in \mathbf{Z}_{+}$,
$$
\left|a_{n}\right| \leqslant|a| \text {. }
$$

Therefore, $\left|a_{n}\right| \leqslant 2\left(n \in \mathbf{Z}_{+}\right)$.
Thus, $M \supseteq\left[-2, \frac{1}{4}\right]$.
In summary, $M=\left[-2, \frac{1}{4}\right]$.