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Hint:

By symmetry, one may assume that the first individual is located on the horizontal radius on the left.

The surface he can see is the portion of the lane cut by the tangents to the tower. The area of this surface, $R(r)$, can be computed with a little bit of trigonometry, as the sum of an annular sector, two right triangles and two segments. We can use a reduced area, i.e. the fraction of the whole ring which is visible.

Then you need to compute the average area of this zone for all positions on the radius. As we assume a model of uniform distribution, the positions must be weighted by the distance to the center, as longer circumferences have higher probabilities (said differently, the element of area in polar coordinates has a factor $r\,dr$).

$$P=\frac{\displaystyle\int_A^{A+B}R(r)\,r\,dr}{\pi((A+B)^2-A^2)}.$$

The aperture of the annular sector is $2\alpha=2\arccos\dfrac Ar$.

Then the equation of a tangent:

$$(-r,0)+t(\sin\alpha,\cos\alpha).$$

The tangency point is given by

$$t_t=r\sin A=\sqrt{r^2-A^2},$$

$$(x_t,y_t)=\frac Ar(-A,\sqrt{r^2-A^2}),$$

and the intersection with the outer circle, by

$$t_i=r\sin\alpha+\sqrt{r^2\sin^2\alpha+r\sin\alpha((A+B)^2-r^2)},$$ $$(x_i,y_i)=(-r,0)+t_i(\sin\alpha,\cos\alpha).$$

A triangle has height $B$ and basis $t_i-t_t$, and a segment has radius $A+B$ and aperture angle

$$\arctan\frac{y_i}{x_i}+\pi-\alpha.$$

Computing the integral looks like a tremendous task.

With $S$ the area of the annulus,

$$SR(r)=\frac S\pi\arccos\frac Ar+B\sqrt{r^2-A^2+\sqrt{r^2-A^2}((A+B)^2-r^2)}+(A+B)^2\left(\beta-\sin\frac{\beta}2\right)$$ where

$$\beta=\arctan\frac{\left(\sqrt{r^2-A^2}+\sqrt{r^2-A^2+\sqrt{r^2-A^2}((A+B)^2-r^2)}\right)\dfrac Ar}{-r+\left(\sqrt{r^2-A^2}+\sqrt{r^2-A^2+\sqrt{r^2-A^2}((A+B)^2-r^2)}\right)\dfrac{\sqrt{r^2-A^2}}r}\\+\pi-\arccos\frac Ar.$$

Sigh.