1. Solve the system

$$
\left\{\begin{array}{l}
3^{2 x-2 y}+2 \cdot 3^{x-y}-3=0 \\
3^{x}+3^{1-y}=4
\end{array}\right.
$$
Solution. Let's introduce the substitution $3^{x}=u, 3^{-y}=v$, where $u, v>0$. The system takes the form

$$
\left\{\begin{array}{l}
u^{2} v^{2}+2 u v-3=0 \\
u+3 v=4
\end{array}\right.
$$

or equivalently

$$
\left\{\begin{array}{l}
(u v)^{2}+2 u v-3=0 \\
u+3 v=4
\end{array}\right.
$$

We solve the first equation as a quadratic in $u v$, yielding two solutions $u v=1$ and $u v=-3$. Since $u, v>0$, we only consider the solution $u v=1$. To find the values of $x, y$, we need to solve the system

$$
\left\{\begin{array}{l}
u v=1 \\
u+3 v=4
\end{array}\right.
$$

The solutions to this system are the solutions to the quadratic equation $v(4-3 v)-1=0$, i.e., $v_{1}=1$ and $v_{2}=\frac{1}{3}$. The corresponding values for $u$ are $u_{1}=1$ and $u_{2}=3$.

Finally, solving the systems

$$
\left\{\begin{array} { l } 
{ 3 ^ { x } = 1 } \\
{ 3 ^ { - y } = 1 }
\end{array} \text { and } \left\{\begin{array}{l}
3^{x}=3 \\
3^{-y}=\frac{1}{3}
\end{array}\right.\right.
$$

yields the solutions to the original system $(x, y) \in\{(0,0),(1,1)\}$.