jmc

Let $y=f(x)$. Then, $f(f(x))=f(y)=3$, so either $2y-5=3$ or $-y+5=3$. If $2y-5=3$, then $y=4$. Note that $4\ge 3$, so $f(4)=3$. If $-y+5=3$, then $y=2$. Note that $2<3$, so $f(2)=3$. So the two possible values of $y$ are 2 and 4.

Now, we solve the equation $f(x)=2$. In this case, either $2x-5=2$ or $-x+5=2$. If $2x-5=2$, then $x=7/2$. Note that $7/2\ge 3$, so $f(7/2)=2$. If $-x+5=2$, then $x=3$. But $3\ge 3$, so $f(3)=2\cdot 3-5=1$, which is not 2.

Next, we solve the equation $f(x)=4$. In this case, either $2x-5=4$ or $-x+5=4$. If $2x-5=4$, then $x=9/2$. Note that $9/2\ge 3$, so $f(9/2)=3$. If $-x+5=4$, then $x=1$. Note that $1<3$, so $f(1)=4$.

Therefore, there are $\boxed{3}$ solutions to $f(f(x))=3$, namely $x=1$, 7/2, and 9/2.