1. In $\triangle A B C$, it is known that $\angle B=2 \angle A, B C=$ $2, A B=2+2 \sqrt{3}$. Then $\angle A=$ $\qquad$ .
$=1.15^{\circ}$.
As shown in Figure 5, draw $CE \perp AB$ at point $E$, extend $AB$ to $D$ such that $BD = BC$, and connect $CD$.
Then $\angle D = \angle BCD = \frac{1}{2} \angle ABC = \angle A$.
Therefore, $AE = DE = \frac{1}{2} AD = \frac{1}{2}(AB + BD) = 2 + \sqrt{3}$,
$BE = AB - AE = (2 + 2\sqrt{3}) - (2 + \sqrt{3}) = \sqrt{3}$.
In the right triangle $\triangle BCE$,
$\cos \angle EBC = \frac{EB}{BC} = \frac{\sqrt{3}}{2} \Rightarrow \angle EBC = 30^{\circ}$.
Thus, $\angle A = \frac{1}{2} \angle ABC = 15^{\circ}$.