1. Given the function $f(x)=[x[x[x]]]$, where $[x]$ denotes the greatest integer not exceeding $x$. If its domain is $[0,4]$, then its range is $\qquad$ .
Ni.1. $\{0,1,8,9,12,13,14,27,28,29,33,34,35, 36,40,41,42,43,64\}$.
(1) When $0 \leqslant x<1$, $[x]=0$, then $f(x)=0$.
(2) When $1 \leqslant x<2$, $[x]=1$, then $f(x)=1$.
(3) When $2 \leqslant x<3$, let $x=2+\alpha, \alpha \in[0,1)$. Then
$$
\begin{array}{l}
f(x)=[(2+\alpha)[(2+\alpha) \times 2]] \\
=[(2+\alpha)(4+[2 \alpha])] \\
=8+[4 \alpha+2[2 \alpha]+\alpha[2 \alpha]] .
\end{array}
$$

If $2 \alpha \in[0,1)$, then
$$
f(x)=8+[4 \alpha], 4 \alpha \in[0,2) \text {. }
$$

Thus, $f(x)$ can take 8,9.
If $2 \alpha \in[1,2)$, then
$$
f(x)=10+[5 \alpha], 5 \alpha \in\left[\frac{5}{2}, 5\right) \text {. }
$$

Thus, $f(x)$ can take $12,13,14$.
(4) When $3 \leqslant x<4$, let $x=3+\beta, \beta \in[0,1)$. Then
$$
\begin{array}{l}
f(x)=[(3+\beta)[(3+\beta) \times 3]] \\
=[(3+\beta)(9+[3 \beta])] \\
=27+[9 \beta+3[3 \beta]+\beta[3 \beta]] .
\end{array}
$$

If $3 \beta \in[0,1)$, then
$$
f(x)=27+[9 \beta], 9 \beta \in[0,3) \text {. }
$$

Thus, $f(x)$ can take $27,28,29$.
If $3 \beta \in[1,2)$, then
$$
f(x)=30+[10 \beta], 10 \beta \in\left[\frac{10}{3}, \frac{20}{3}\right) \text {. }
$$

Thus, $f(x)$ can take $33,34,35,36$.
If $3 \beta \in[2,3)$, then
$$
f(x)=33+[11 \beta], 11 \beta \in\left[\frac{22}{3}, 11\right) \text {. }
$$

Thus, $f(x)$ can take $40,41,42,43$.
Also, $f(4)=64$. In summary, this gives the range of $f(x)$.