3. Rewrite the system as

$$
\left\{\begin{array}{l}
|x-1|-y=1-a^{4}-a^{4}(x-1)^{4} \\
(x-1)^{2}+y^{2}=1
\end{array}\right.
$$

and make the substitution $\mathrm{t}=\mathrm{x}-1$.

We get

$$
\left\{\begin{array}{l}
|t|-y=1-a^{4}-a^{4} t^{4} \\
t^{2}+y^{2}=1
\end{array}\right.
$$

We see that if (t, y) is a solution, then (-t, y) is also a solution. If the solution is unique, then $t=0$, so

$$
\left\{\begin{array}{l}
y=a^{4}-1 \\
y^{2}=1
\end{array}\right.
$$

For $\mathrm{y}=-1 \mathrm{a}=0$, and for $\mathrm{y}=1 \mathrm{a}^{4}=2$.

Let $\mathrm{a}=0$, then the system becomes

$$
\left\{\begin{array}{l}
|t|-y=1 \\
t^{2}+y^{2}=1
\end{array}\right.
$$

There are more than one solution, for example, $(0,-1),(1,0)$ (the first value in the pair is $t$).

Let $\mathrm{a}^{4}=2$, then the system has the form

$$
\left\{\begin{array}{l}
|t|-y=-1-2 t^{4} \\
t^{2}+y^{2}=1,
\end{array},\left\{\begin{array}{l}
y=1+|t|+2 t^{4} \\
y^{2}=1-t^{2} .
\end{array}\right.\right.
$$

From the first equation, it follows that $\mathrm{y} \geq 1$, and $\mathrm{y}=1$ only when $\mathrm{t}=0$.

In the second equation, for $\mathrm{t} \neq 1 \mathrm{y}<1$. Therefore, for $\mathrm{t} \neq 0$ the system has no solutions, and $(0,1)$ is the only solution when $\mathrm{a}^{4}=2$.
Answer: $\pm \sqrt[4]{2}$.