$\square$ Answer: $A_{13}$.

Fig. 2. Note that if $180^{\circ}-\angle A_{i} A_{i-1} A_{0}<90^{\circ}$, then $A_{i+1}$ is farther from $A_{0}$ than $A_{i-1}$, and the triangle $A_{i-1} A_{i} A_{i+1}$ is isosceles, with

$$
180^{\circ}-\angle A_{i+1} A_{i} A_{0}=\angle A_{i} A_{i+1} A_{0}+7^{\circ}=\left(180^{\circ}-\angle A_{i} A_{i-1} A_{0}\right)+7^{\circ}
$$

Therefore, the value $180^{\circ}-\angle A_{i} A_{i-1} A_{0}$ increases by $7^{\circ}$ with each iteration (and at $i=2$ it is $14^{\circ}$). It can increase until it becomes greater than $90^{\circ}$ - after which the points will start to get closer to $A_{0}$. This will happen at $i=13\left(7^{\circ} \cdot 13=91^{\circ}\right)$. It remains to note that $A_{0} A_{12}<A_{0} A_{13}$. Indeed, this is equivalent to $84^{\circ}=\angle A_{0} A_{13} A_{12}<\angle A_{0} A_{12} A_{13}=$ $89^{\circ}$.

(Old Kangaroo competition)
5. From point $A_{0}$, a black and a red ray are drawn at an angle of $7^{\circ}$, after which a broken line $A_{0} A_{1} \ldots A_{20}$ (possibly self-intersecting, but all vertices are distinct) is constructed, where all segments have a length of 1, all even vertices lie on the black ray, and all odd vertices lie on the red ray. Which vertex is the farthest from vertex $A_{0}$?