3. The only such number is $n=1$.

Let us prove by induction on $a \in \mathbb{N}_{0}$ the following statement: if the number $n=\overline{t_{k} t_{k-1} \ldots t_{1} t_{0}}$ satisfies the given conditions, for $k \geqslant a$, then $t_{0}=1$ and $t_{1}=t_{2}=\cdots=t_{a}=0$. This statement directly solves the problem, since for $k=a \geqslant 1$ we have a contradiction.

First, take $a=0$ and $n=\overline{t_{k} t_{k-1} \ldots t_{1} t_{0}}$. Clearly, $t_{0} \neq 0,2,8$ (the first case is ruled out because then $t_{k}=0$, and the remaining two because then $n^{2}$ would end in the digit 4). Assume $t_{0}=5$. Then $t_{k}=2$, so we have $2 \cdot 10^{k}<n<3 \cdot 10^{k}$. From this, $n^{2}$ starts with one of the digits $4,5, \ldots, 8$ - but this is impossible, since $n^{2}$ ends in the digit 5, so it would have to start with the digit 2. This proves the base case of the induction.

Now assume the statement holds for $a-1$, and prove it for $a$. Let $n=\overline{t_{k} t_{k-1} \ldots t_{1} t_{0}}$ satisfy the given conditions, and $k \geqslant a$. By the induction hypothesis, we have $t_{0}=1$ and $t_{1}=t_{2}=\cdots=t_{a-1}=0$. Thus, $n=10^{a+1} x+10^{a} t_{a}+1$. For $t_{a}=2$, we would have $n^{2} \equiv 4 \cdot 10^{a}+1\left(\bmod 10^{a+1}\right)$, i.e., $n^{2}$ would have the digit 4, a contradiction. Similarly, for $t_{a}=8$, we would have $n^{2} \equiv 6 \cdot 10^{a}+1\left(\bmod 10^{a+1}\right)$, again a contradiction. Now assume $t_{a}=1$. Then we have $n^{2}=\ldots 2 \underbrace{00 \ldots 00}_{a-1} 1$,

$$
\begin{gathered}
\text { so it follows that } n^{2}=1 \underbrace{00 \ldots 00}_{a-1} 5 \ldots \text { On the other hand, we have } n=1 \underbrace{00 \ldots 00}_{a-1} 1 \ldots, \text { i.e., } \\
10^{k}+10^{k-a}<n<10^{k}+2 \cdot 10^{k-a}, \text { and from this } \\
10^{2 k}+2 \cdot 10^{2 k-a}+10^{2 k-2 a}<n<10^{2 k}+4 \cdot 10^{2 k-a}+4 \cdot 10^{2 k-2 a},
\end{gathered}
$$

a contradiction. Similarly, for $t_{a}=5$ we get $n^{2}=1 \underbrace{00 \ldots 00}_{a} \ldots$ but

$$
10^{2 k}+4 \cdot 10^{2 k-a}+4 \cdot 10^{2 k-2 a}<n^{2}<10^{2 k}+6 \cdot 10^{2 k-a}+9 \cdot 10^{2 k-2 a}
$$

and again a contradiction. The only possibility left is $t_{a}=0$, which completes the proof.
3. Let the function odraz map the digits $0,1,2,5,8$ to the digits $0,1,5,2,8$, respectively. A natural number $n=\overline{t_{k} t_{k-1} \cdots t_{1} t_{0}}$ is called reflectable if all its digits are from the set $\{0,1,2,5,8\}$ and $t_{0} \neq 0$, and we define

$$
\operatorname{odraz}(n)=\overline{\operatorname{odraz}\left(t_{0}\right) \operatorname{odraz}\left(t_{1}\right) \cdots \operatorname{odraz}\left(t_{k-1}\right) \operatorname{odraz}\left(t_{k}\right)}
$$

(in other words, the function odraz represents the reflection in a mirror of the number on a calculator screen). Find all natural numbers $n$ with the following properties:

$1^{\circ} n$ is reflectable and $\operatorname{odraz}(n)=n$;

$2^{\circ} n^{2}$ is reflectable and $\operatorname{odraz}\left(n^{2}\right)=n^{2}$.