(13) Find the intersection points of the curves $y=2-\sqrt{1-4^{x}}$ and $y=a \cdot 2^{x}(a \in \mathbf{R})$.
(13 Let $t=2^{x}$, the problem is transformed into the line $y=a t$ and the arc $y=2-\sqrt{1-t^{2}}(00$ and $y \leqslant$ 2. That is, the line $y=a t$ and the circle $t^{2}+(y-2)^{2}=1$ intersect under the conditions $a>0$ and $y \leqslant 2$, with the intersection point in the first quadrant, hence $0\sqrt{3}$, equation (1) has two real roots $t_{1}=\frac{2 a-\sqrt{a^{2}-3}}{a^{2}+1}, t_{2}=$ $\frac{2 a+\sqrt{a^{2}-3}}{a^{2}+1}$. The line intersects the circle at two points $\left(t_{1}, y_{1}\right),\left(t_{2}, y_{2}\right)$. At this time
$$
\begin{array}{l}
y_{1}=a t_{1}=2-\frac{2+a \sqrt{a^{2}-3}}{a^{2}+1} \leqslant 2, \\
y_{2}=a t_{2}=2+\frac{a \sqrt{a^{2}-3}-2}{a^{2}+1} .
\end{array}
$$

When $\sqrt{3}2$.
In summary, when $a2$, the two curves intersect at one point $\left(\log _{2} \frac{2 a-\sqrt{a^{2}-3}}{a^{2}+1}\right.$,
$$
\left.2-\frac{2+a \sqrt{a^{2}-3}}{a^{2}+1}\right) .
$$