I am wondering if the interpretation used by caligirl11 and @joriki is the one that was intended by whomsoever devised the problem.
What is the probability that she wins the game on the kth point played for k=4,5,6,...
I suggest that the $k$-th point played is not the same as the $k$-th point won by Anne.
Ok so I started with k=4 P(win on 4th point) can happen 3 ways 4-0,4-1,4-2 =p^4(1 + (4 choose 1)q +(5 choose 2)q^2)
I would say here that Anne wins the game on the $4$th point played with probability $p^4$; the probability that she wins on the $5$th point played is $\binom{4}{1}p^4q$; and the probability that she wins on the $6$th point played is $\binom{5}{2}p^4q^2$. The corresponding win probabilities for Betty are obtained by interchanging $p$ and $q$ in the previous sentence.
If $6$ points have been played and the game is not over, the score must be $3$-$3$ (deuce) and this event has probability $\binom{6}{3}p^3q^3 = 20p^3q^3$. The game can now end only when an even number of additional points have been played, and the probability of Anne winning on the $k$-th point played (where k = 6+2n, n>0) is $\binom{6}{3}p^3q^3\cdot p^2(2pq)^{n-1}$. To get the probability that Anne wins the game, sum the probabilities that she wins in $4$, $5$, $6$, $8$, $10, \ldots$ points; a geometric series is involved.
If $f(p)$ denotes the probability that Anne wins the game, then $f(0.5) = 0.5$ but $\left . \frac{d}{dp}f(p)\right\vert_{p = 0.5} = 2.5$ so that $f(0.5 + \epsilon) = 0.5 + 2.5\epsilon + \cdots$, that is, a small difference $p-q = 2\epsilon$ in the point win probabilities for Anne and Betty is amplified by the rules of tennis into a greater difference in the game win probabilities. See "The Drunken Tennis Player" in Ian Stewart's Game, Set, and Math, Penguin Books, 1981.