Let the center of $s_i$ be $O_i$. $\angle O_1AP_1 + \angle O_2AO_1 + \angle P_2AO_2 = 0$, so we need to maximize the product of cosines of two angles which sum to a constant angle. $\cos x \cos (A - x) = \frac{\cos A + \cos (2x - A)}{2}$, so the minimum is achieved when $x = \frac{A}{2}$. This corresponds to either of the angle bisectors of $\angle O_1AO_2$ (by similar triangles it is easy to see that they give the same answer, though in terms of directed lengths, they have opposite signs).

A bit similar to the above one. Let $\angle P_1BA= \varphi_1$ and $\angle P_2BA=\varphi_2$. It's easy to see that $AP_1=2R_1sin \varphi_1, AP_2=2R_2sin \varphi_2$ and that $\varphi_1+\varphi_2=const$, so we are to maximize $sin \varphi_1. sin \varphi_2=\frac{1}{2}(cos(\varphi_1-\varphi_2)-cos(\varphi_1+\varphi_2))$, so we have to maximize $cos(\varphi_1-\varphi_2)$ and since $cos(x)$ is decreasing for $x$ in $[0, \pi]$, $\varphi_1=\varphi_2$. Now take random line intersecting the two circles at $P_1', P_2'$, let $\angle P_1'BP_2'=2\varphi$, construct its angle bisector to measure $\varphi$, and now take $P_1$ on $\omega_1$, such that $\angle P_1BA=\varphi$ let $P_1A$ meet $\omega_2$ at $P_2$, so the construction is finished.

Well we need to draw the angle bisector of $\angle O_1AO_2$(= x) if $x$ is acute to get the points $P_1$ and $P_2$ as intersections of this bisector with the circles. If $x$ is obtuse draw the perpendicular to the above mentioned bisector.