412. Solve the system of equations:

$$
\left\{\begin{array}{l}
x+y+\sqrt{x^{2}-y^{2}}=12 \\
y \sqrt{x^{2}-y^{2}}=12
\end{array}\right.
$$
$\triangle$ Let $\sqrt{x^{2}-y^{2}}=z$, where $z \geq 0$. We obtain the system of rational equations

$$
\left\{\begin{array}{l}
x+y+z=12 \\
y z=12 \\
x^{2}-y^{2}=z^{2}
\end{array}\right.
$$

Since

$$
x^{2}-y^{2}=z^{2}, \quad x+y=12-z
$$

then $x-y=\frac{z^{2}}{12-z}$. From the system of equations

$$
x+y=12-z, \quad x-y=\frac{z^{2}}{12-z}
$$

we express $x$ and $y$ in terms of $z$:

$$
x=\frac{z^{2}-12 z+72}{12-z}, \quad y=\frac{72-12 z}{12-z}
$$

Substitute the expression for $y$ into the second equation of the system of three equations:

$$
\frac{72-12 z}{12-z} z=12, \quad(6-z) z=12-z, \quad z^{2}-7 z+12=0
$$

From this, $z_{1}=4, z_{2}=3$. Now we easily find $x$ and $y$.

Answer: $(5 ; 3),(5 ; 4)$.