1. Yes. If there is another one, it differs from $L'$ by a line bundle $M$ with $M^{2}\cong \mathcal{O}_Y$. Consider the resolution $\pi :\hat{Y}\rightarrow Y$ obtained by blowing up the double points $p_1,\ldots ,p_{16}$. Since $\hat{Y}$ is simply connected, we have $\pi^* M\cong \mathcal{O}_{\hat{Y}}\ $.

Thus $M_{|Y\smallsetminus \{p_i\}}$ is trivial, and this implies that $M$ is trivial.

2. Yes by 1), since $\pi ^*\mathcal{O}_Y(C')=\mathcal{O}_X(C)=L^2$.

1. Yes. If there is another one, it differs from $L'$ by a line bundle $M$ with $M^{2}\cong \mathcal{O}_Y$. Consider the resolution $\pi :\hat{Y}\rightarrow Y$ obtained by blowing up the double points $p_1,\ldots ,p_{16}$. Since $\hat{Y}$ is simply connected, we have $\pi^* M\cong \mathcal{O}_{\hat{Y}}\ $.

Thus $M_{|Y\smallsetminus \{p_i\}}$ is trivial, and this implies that $M$ is trivial.

2. Yes by 1), since the pull-back of $\mathcal{O}_Y(C')$ to $X$ is $\mathcal{O}_X(C)=L^2$.