Let the points be $P_1$, $P_2$, and $P_3$. Letting $P_1$ be arbitrary, we win if $P_2$ is the point diametrically opposite to $P_1$, probability $1/2009$. Otherwise, with probability $2008/2009$, we still win if $P_3$ is diametrically opposite either $P_1$ or $P_2$, which it is with probability $2/2008$. So the probability of a win is $1/2009 + (2008/2009)(2/2008) = 3/2009$.
You can also get this as follows: The number of possible wins is the number of diameters times the number of remaining points, or 1005*2008. The number of possible triples is $2010\choose 3$. Dividing the first by the second gives $3/2009$.
Even easier: The probability that a given pair of points lie on a diameter is $1/2009$. With three points, you have three pairs, hence $3/2009$. Here it's OK to add probabilities because it's not possible to have overlap; if one of the pairs lies on a diameter, no other pair can lie on a diameter.
BTW, I do not agree with the comment thread that says we must consider the possibility that the points are not distinct. Sometimes there can be an ambiguity, but in common language "choose three points from 2010 points" means "without replacement". For example, I am not incorrect in saying that $n\choose r$ is the number of ways of choosing $r$ objects out of a set of $n$. Although I admit that if I were grading the test and this issue were pointed out to me, I might accept the "with replacement" answer as well.
The only other alternative answer I might accept is 1, i.e., that the points are certain to lie on a right triangle. After all, it is Ford Prefect, so I have no way of knowing whether the Infinite Improbability Drive has been activated.
The probability that point $A$ is at a distance from the center in $[a,a+da]$ is $2\pi a\,da/\pi=2a\,da$.
The probability that, fixed $A$ as above, $\angle ABO>90°$, is the same as the probability that $B$ lies inside a circle of diameter $OA$, that is $\pi(a/2)^2/\pi=a^2/4$.
Hence the overall probability that $\angle ABO>90°$ is:
$$
p(\angle ABO>90°)=\int_0^1 {a^2\over4}\cdot 2a\,da={1\over8}.
$$
The probability that triangle $ABO$ is obtuse is then:
$$
p(\angle ABO>90°)+p(\angle BAO>90°)+p(\angle AOB>90°)
={1\over8}+{1\over8}+{1\over2}={3\over4}.
$$
EDIT.
For a 3D sphere one can repeat the same argument, obtaining:
$$
p(\angle ABO>90°)=\int_0^1 {a^3\over8}\cdot 3a^2\,da={1\over16}.
$$
Best Answer
The number $576\choose 2$ comes from the number of ways to choose the other two vertices of the obtuse-angled triangle, once you've chosen the vertex with the obtuse angle.
With the point chosen for your obtuse angle, label all points with numbers $-576$ to $576$, e.g. in clockwise order, so that your chosen point is labelled as $0$. We need to pick two of those - one at each side of the given point $0$ (i.e. one negative and one positive), so that the angle is actually obtuse. This means that the "distance" of the second and the third point is at most $576$.
One can see that, if you pick $-576$, you have no choices for the third point at all (as you cannot choose $0$ again, and any positive point is "further away"). If you pick $-575$ you can only pick $1$. If you pick $-574$, you can also pick $1$ or $2$ etc. until, if you pick $-1$, you can pick any of the the points $1,2,3,\ldots,575$.
So the total number of ways to pick the "other" two points is $1+2+3+\ldots+575={576\choose 2}$.
I suggest you test this logic on something simpler, i.e. instead of $1153$ points choose any small odd number of points (say $5$ or $7$) and convince yourselves that the same logic works. E.g. for $5$ points there is only one way to choose the other two ponts, for $7$ points there are $1+2=3$ ways etc.
Update: I can see from @Golden_Ratio's answer that the value $576\choose 2$ pops out directly if, instead of the obtuse angle vertex we choose the acute angle vertex (the first in, say, clockwise, order). Still, I will keep this answer as a potentially different (less efficient for sure!) solution.