Solution:

Answer: $\sqrt{3}/2$

Let the points be $A$, $B$, $C$, $X$, $Y$ so that $X$ and $Y$ are on opposite sides of the plane defined by triangle $ABC$. The volume is $1/3$ the product of the area of $ABC$ and the sum of the distances from $X$ and $Y$ to the plane defined by $ABC$.

The area of $ABC$ is maximized when the plane containing it intersects the sphere in the largest possible cross section, and $ABC$ is equilateral: this gives an area of $3\sqrt{3}/4$.

Then, the sum of the distances from $X$ and $Y$ to the plane of $ABC$ is at most $2$. This is clearly obtainable when $A$, $B$, and $C$ form an equilateral triangle circumscribed by the equator of the sphere, and $X$ and $Y$ are at the poles, and we get a volume of $\sqrt{3}/2$.