3. Set $A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}y=\sqrt{1-x}, \\ y=1-x^{2}\end{array}\right\}\right.\right.$ has the number of subsets as $\qquad$
3. 8 .

Notice that the number of elements in set $A$ is the number of solutions to the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2}
\end{array}\right.
$$

Substituting equation (1) into equation (2) and simplifying, we get
$$
x(x-1)\left(x^{2}+x-1\right)=0 \text {. }
$$

Solving this, we get $x_{1}=0, x_{2}=1$,
$$
x_{3}=\frac{-1+\sqrt{5}}{2}, x_{4}=\frac{-1-\sqrt{5}}{2} \text {. }
$$

Noting that $\left|x_{4}\right|>1$, we discard it.
Therefore, the system of equations has three solutions
$$
(0,1),(1,0),\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right) \text {. }
$$

Since set $A$ has 3 elements, set $A$ has $2^{3}=8$ subsets.