jmc

Since $\sin E=1$, we have $\angle E=90^{\circ }$, so our triangle is as shown below:



Since $\sin D=\frac{5}{13}$, we have $\frac{EF}{DF}=\frac{5}{13}$, so $\cos F=\frac{EF}{DF}=\frac{5}{13}$. Since ${\sin }^{2}F+{\cos }^{2}F=1$, and $\angle F$ is acute (so $\sin F$ is positive), we have $$\sin F=\sqrt{1-{\cos }^{2}F}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\boxed{\frac{12}{13}}.$$We also could have noticed that since $\frac{EF}{DF}=\frac{5}{13}$, we have $EF=5x$ and $DF=13x$ for some value of $x$. Then, from the $\{5,12,13\}$ Pythagorean triple, we see that $DE=12x$, so $\sin F=\frac{DE}{DF}=\frac{12}{13}$.