5. Specify all values of $a$ for which the system of equations $3(x-a)^{2}+y=2-a$, $y^{2}+\left(\frac{x-2}{|x|-2}\right)^{2}=1$ has at least one solution, and solve it for each $a$.
Solution. Transform the second equation: $y^{2}=\frac{x^{2}-4|x|+4-x^{2}+4 x-4}{(|x|-2)^{2}} ; y^{2}=-\frac{4(|x|-x)}{(|x|-2)^{2}}$.

For $x<0$ the equation has no solutions, as the right side is less than zero; for $x \geq 0$ it is equivalent to the system: $x \geq 0, x \neq 2, y=0$. Substituting $y=0$ into the first equation, we get: $3(x-a)^{2}=2-a$, or $3 x^{2}-6 a x+3 a^{2}+a-2=0(*)$, for which $D / 4=9 a^{2}-9 a^{2}-3 a+6=6-3 a$.

The root $x=2$ of the quadratic equation can occur when $3(2-a)^{2}=2-a$, i.e., if 1) $a=2$, the equation takes the form $(x-2)^{2}=0$, then this root is the only one and the given system has no solutions, or 2) $3(2-a)=1$, i.e., $a=5 / 3$, then for $x$ we get the equation $3(x-5 / 3)^{2}=2-5 / 3, x=5 / 3 \pm 1 / 3$, which, besides the extraneous root $x_{1}=2$, has another root $x_{2}=4 / 3$, satisfying the conditions, and the given system has a unique solution $(4 / 3 ; 0)$.

Consider the other cases when the system will have a unique solution.