Probability that the player wins a pass line bet with a 4 on the first roll

Yes, your solution is correct. Here's another approach. Let $p$ be the probability that, after the first roll of $4$, the second $4$ appears before the first $7$. Then by conditioning on whether the next roll is $4$, $7$, or something else, we find that \begin{align} p &= \mathbb{P}[\text{$4$ before $7$}] \\ &= \mathbb{P}[4]\cdot \mathbb{P}[\text{$4$ before $7$}\mid4] + \mathbb{P}[7]\cdot \mathbb{P}[\text{$4$ before $7$}\mid7] + \mathbb{P}[\text{other}]\cdot \mathbb{P}[\text{$4$ before $7$}\mid \text{other}] \\ &= \frac{3}{36}\cdot 1 + \frac{6}{36}\cdot 0 + \frac{27}{36}\cdot p \end{align} yielding $p=\frac{3/36}{9/36}=\frac{1}{3}$. Hence the desired probability is $$\mathbb{P}[4]\cdot p=\frac{3}{36}\cdot\frac{1}{3}=\frac{1}{36}.$$