= Solution
Scale the disk to the <unit disc>, so that the starting point is $a=1/r\in(0,1)$. Use the <Möbius transformation>
$$
\psi_a(w)=\frac{w-a}{1-aw},
$$
which maps the disk onto itself, sends $a$ to $0$, and sends $1$ to $1$. By <conformal invariance of planar Brownian motion>, the exit image is the circular exit of a <Brownian motion> starting at $0$. <Rotational invariance of planar Brownian motion> makes that exit uniform in angle.
The endpoints of the right semicircle satisfy
$$
\psi_a(i)=\frac{-2a+i(1-a^2)}{1+a^2}=e^{i\varphi},\qquad
\psi_a(-i)=e^{-i\varphi},\qquad
\varphi=\frac\pi2+2\arctan a.
$$
Its image is the arc through $1$ between these points, of angular length $2\varphi$. Thus the <Möbius calculation of circular Brownian exit> gives
$$
\mathbb P_1(E_+)=\frac{2\varphi}{2\pi}=\frac12+\frac2\pi\arctan(1/r).
$$
Since $E_+$ and $E_-$ partition the exit almost surely, part (b) yields
$$
\boxed{\mathbb P_1(T(r)<S)=\frac4\pi\arctan(1/r)
=\frac2\pi\arctan\!\left(\frac{2r}{r^2-1}\right).}
$$
The last equality uses $2\arctan(1/r)\in(0,\pi/2)$, so the double-angle tangent identity uses the stated principal branch. As a check, the boundary angle derivative of $\psi_a$ is $(1-a^2)/|e^{i\theta}-a|^2$; integrating this circular exit density over the right semicircle gives exactly the integral supplied in the question.
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