= Solution
Let $\Theta=\angle AOB$ and let the <sphere> have radius $R$. By rotational invariance, condition on $A$ and place it at the north pole. A band at polar angle $\theta$ has <surface area> $2\pi R^2\sin\theta\,d\theta$, while the sphere has area $4\pi R^2$. Thus the <central angle between independent uniform sphere points> has density
$$
\boxed{f_\Theta(\theta)=\frac12\sin\theta\quad(0<\theta<\pi),}
$$
with density zero outside that interval.
Condition now on $A,B$ with their angle fixed at $\theta$. For $\angle AOC$ and $\angle BOC$ to be acute, the unit position <vector> of $C$ must have positive <dot products> with both unit vectors of $A$ and $B$. Hence $C$ must lie in the intersection of the two corresponding open <hemispheres>. The great circles bounding them meet at the two antipodes perpendicular to the plane of $A,B$. Their common hemisphere region is a <spherical lune> of opening $\pi-\theta$: when $A=B$ it is a hemisphere of opening $\pi$, and as the two normals separate, that opening decreases by $\theta$. More explicitly, take the common boundary antipodes as poles. The longitude conditions each allow an interval of length $\pi$, with their centres separated by $\theta$, so the intersection has longitude width $\pi-\theta$.
Integrating the spherical area element $R^2\sin\varphi\,d\varphi\,d\lambda$ over that lune gives area $2R^2(\pi-\theta)$. Thus the conditional probability that the two angles involving $C$ are acute is $(\pi-\theta)/(2\pi)$. The remaining angle $\Theta$ must itself be acute, so the desired probability is
$$
P=\int_0^{\pi/2}\frac{\pi-\theta}{2\pi}\frac{\sin\theta}{2}\,d\theta.
$$
Integration by parts gives $\int_0^{\pi/2}\theta\sin\theta\,d\theta=1$, while $\int_0^{\pi/2}\sin\theta\,d\theta=1$. Therefore
$$
\boxed{P=\frac{\pi-1}{4\pi}.}
$$
This calculates <pairwise acute central angles of three uniform sphere points> by conditional area, rather than incorrectly treating the three angle events as independent. Boundary right angles have probability zero, so choosing open rather than closed hemispheres does not change the result.
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