jmc

For the ellipse $\frac{x^2}{49}+\frac{y^2}{33}=1,$ $a=7$ and $b=\sqrt{33},$ so $$c^2=a^2-b^2=49-33=16.$$Then $c=4,$ so $F_1=(4,0)$ and $F_2=(-4,0).$

Since $Q$ lies on the ellipse, $F_{1}Q+F_{2}Q=2a=14.$ Then $$F_{2}P+PQ+F_{1}Q=14,$$so $PQ+F_{1}Q=14-F_{2}P.$ Thus, we want to minimize $F_{2}P.$

Let $O=(0,3),$ the center of the circle $x^2+(y-3{)}^2=4.$ Since $P$ lies on this circle, $OP=2.$ By the Triangle Inequality, $$F_{2}P+PO\ge F_{2}O,$$so $F_{2}P\ge F_{2}O-PO=5-2=3.$ Equality occurs when $P$ lies on line segment $\overline{F_{2}O}.$



Therefore, the maximum value of $PQ+F_{1}Q$ is $14-3=\boxed{11}.$