Let $p$ be the probability that Betsy wins the game. We check some cases. If Betsy flips tails on her first move, then she wins and the game ends; this happens with probability $2/5$. Otherwise, Katie gets a chance to play; this happens with probability $3/5$. If Katie gets a chance to play, she wins with probability $2/5$, or resets the game with probability $3/5$. Therefore, $$p = \frac{2}{5} \cdot 1 + \frac{3}{5} \left(\frac{2}{5} \cdot 0 + \frac{3}{5} p \right) \implies p = \frac{5}{8}$$
Alternatively, the probability that Betsy wins equals the probability she wins on her first move, plus the probability she wins on her second move, plus ...
The probability Betsy wins on her $k$th move is $$\left(\frac{3}{5} \right)^{2(k-1)} \cdot \frac{2}{5}$$ since every flip before her winning move must have been a heads. So the answer is $$\frac{2}{5} \cdot \sum_{k=1}^{\infty} \left(\frac{3}{5} \right)^{2(k-1)} = \frac{5}{8}$$
Note that both of these solutions are essentially the same.