1. Find the focus $S$ of the hyperbola

Given hyperbola: $\frac{x^2}{3}-\frac{y^2}{5}=1$

This is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $a^2=3,\quad b^2=5.$

For such a hyperbola, $c^2=a^2+b^2=3+5=8 \implies c=2\sqrt{2}.$

So the focus on the positive $x$-axis is $S=(2\sqrt{2},0).$

2. Use the fact that $SAB$ is a diameter of the circle

The circle has centre at $A=(\sqrt{6},\sqrt{5}).$

If $SAB$ is a diameter, then $A$ is the midpoint of $SB$. So if $B=(x_B,y_B),$ then $\left(\frac{2\sqrt{2}+x_B}{2},\frac{0+y_B}{2}\right)=(\sqrt{6},\sqrt{5}).$

Hence, $x_B=2\sqrt{6}-2\sqrt{2},\qquad y_B=2\sqrt{5}.$

Therefore, $B=(2\sqrt{6}-2\sqrt{2},\,2\sqrt{5}).$

3. Find the area of triangle $OSB$

We have $O=(0,0),\quad S=(2\sqrt{2},0),\quad B=(2\sqrt{6}-2\sqrt{2},2\sqrt{5}).$

Since $O$ and $S$ lie on the $x$-axis, base $OS$ is $OS=2\sqrt{2}.$

The perpendicular distance of $B$ from the $x$-axis is $2\sqrt{5}.$

So area of triangle $OSB$ is