Solution:
a) $\sqrt{3 \sqrt{2 \sqrt{n}}}=\sqrt{\sqrt{\sqrt{3^{4} \cdot 2^{2} \cdot n}}} \in \mathbb{N}$ $1 p$

The number obtained can also be written in the form $\sqrt[8]{3^{4} \cdot 2^{2} \cdot n} \in \mathbb{N}$

We get $n=3^{4a} \cdot 2^{8b-2}$, where $a$ is an odd natural number and $b \in \mathbb{N}^{*}$..... $1 p$

For $a=b=1$ we get $n=3^{4} \cdot 2^{6}=81 \cdot 64=5184$ - solution. $.0,5 p$

If $a=1, b=2$ we get $n=3^{4} \cdot 2^{14}>\overline{xyzt}$

If $a=3, b=1$ we get $n=3^{12} \cdot 2^{6}>\overline{xyzt}$

Therefore, the only solution is $n=5184$.

b) Let $x=\sqrt{2}^{\sqrt{3}}$ and $y=\sqrt{3}^{\sqrt{2}}$, then we have:

$$
\begin{aligned}
& x^{2}=\left(\sqrt{2}^{\sqrt{3}}\right)^{2}=\sqrt{2}^{2 \sqrt{3}}=2^{\sqrt{3}} \text { and } y^{2}=\left(\sqrt{3}^{\sqrt{2}}\right)^{2}=\left(\sqrt{3}^{2}\right)^{\sqrt{2}}=3^{\sqrt{2}} \\
& \left(x^{2}\right)^{\sqrt{2}}=\left(2^{\sqrt{3}}\right)^{\sqrt{2}}=2^{\sqrt{6}}<2^{3}=8 \text { and }\left(y^{2}\right)^{\sqrt{2}}=\left(3^{\sqrt{2}}\right)^{\sqrt{2}}=3^{2}=9 \ldots
\end{aligned}
$$

We obtain $x^{2}<y^{2}$ and since $x, y$ are positive numbers, it follows that $x<y$.
Problem 1. a) Determine the natural numbers $n$, with four digits, such that $\sqrt{3 \sqrt{2 \sqrt{n}}} \in \mathbb{N}$.

b) Compare the numbers $\sqrt{2}^{\sqrt{3}}$ and $\sqrt{3}^{\sqrt{2}}$.