If \[f(x) =
\begin{cases}
2x-5 &\quad \text{if } x \ge 3, \\
-x + 5 &\quad \text{if } x < 3,
\end{cases}
\]then for how many values of $x$ is $f(f(x)) = 3$?
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.