Solution. We have: $\Delta=4-b^{2}$ and $\Delta_{x}=2 a c^{2}+2 c-b c+b$.

If $\Delta \neq 0$, i.e., $b \neq \pm 2$, then the system has a unique solution for any $a$ and $c$.

If $\Delta=0$, then for the system to have a solution, $\Delta_{x}$ must also be $0$. For $b=2$, we get $a c^{2}=-1$, and this equation has a solution for $c$ for any $a \in(-\infty, 0)$. For $b=-2$, we get the equation $a c^{2}+2 c-1=0$, which has a solution for $c$ for any $a \in[-1,+\infty)$.

Therefore, the given system has at least one solution for any $b$ and some $c$ if and only if $a \in[-1,0]$.
3. For which values of $a$, for any $b$ there exists at least one $c$, such that the system

$$
\left\{\begin{array}{l}
2 x+b y=a c^{2}+c \\
b x+2 y=c-1
\end{array}\right.
$$

has at least one solution?