2. Solve the equation $\sqrt{\sqrt{x+2}+\sqrt{x-2}}=2 \sqrt{\sqrt{x+2}-\sqrt{x-2}}+\sqrt{2}$.

Answer: $\frac{257}{16}$.
Solution. Let $t=\sqrt{x+2}+\sqrt{x-2}$, then $\sqrt{\sqrt{x+2}-\sqrt{x-2}}=\frac{2}{\sqrt{t}}$, and the given equation takes the form $\sqrt{t}=\frac{4}{\sqrt{t}}+\sqrt{2}$, or $t-\sqrt{2 t}-4=0$. Solving this equation as a quadratic in $\sqrt{t}$, we find $\sqrt{t}=-\sqrt{2}$ or $\sqrt{t}=2 \sqrt{2}$. The first equation has no solutions, while from the second, returning to the variable $x$, we get

$$
\sqrt{x+2}+\sqrt{x-2}=8 \quad \Leftrightarrow \quad 2 x+2 \sqrt{x^{2}-4}=64 \quad \Leftrightarrow \quad\left\{\begin{array}{l}
x^{2}-4=(32-x)^{2}, \quad \Leftrightarrow \quad x=\frac{257}{16} .
\end{array}\right.
$$

Answer to option 2: $\frac{577}{16} . \quad$ Answer to option 3: $\frac{1025}{16} . \quad$ Answer to option $4: \frac{745}{36}$.