Use the cosine of a sum and cosine of a difference identities to find $\cos (s+t)$ and $\cos (s-t)$.
$\sin \mathrm{s}=\frac{5}{13}$ and $\sin \mathrm{t}=-\frac{3}{5}$, sin quadrant $\mathrm{I}$ and $\mathrm{t}$ in quadrant III
\[
\cos (s+t)=
\]
(Use fractions or pi for any numbers in the expression.)

Step 1 ：We are given that \(\sin s = \frac{5}{13}\) and \(\sin t = -\frac{3}{5}\), and that s is in quadrant I and t is in quadrant III.

Step 2 ：Since s is in quadrant I, both \(\sin s\) and \(\cos s\) are positive. Since t is in quadrant III, both \(\sin t\) and \(\cos t\) are negative.

Step 3 ：We can use the Pythagorean identity \(\cos^2 x = 1 - \sin^2 x\) to find \(\cos s\) and \(\cos t\).

Step 4 ：Then we can substitute these values into the identities to find \(\cos (s+t)\) and \(\cos (s-t)\).

Step 5 ：\(\cos s = \sqrt{1 - \sin^2 s} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}\)

Step 6 ：\(\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = -\frac{4}{5}\)

Step 7 ：Substitute these values into the identities, we get \(\cos (s+t) = \cos s \cos t - \sin s \sin t = \frac{12}{13} \times -\frac{4}{5} - \frac{5}{13} \times -\frac{3}{5} = -\frac{33}{65}\)

Step 8 ：And \(\cos (s-t) = \cos s \cos t + \sin s \sin t = \frac{12}{13} \times -\frac{4}{5} + \frac{5}{13} \times \frac{3}{5} = -\frac{63}{65}\)

Step 9 ：Final Answer: The cosine of the sum of s and t is \(\boxed{-\frac{33}{65}}\) and the cosine of the difference of s and t is \(\boxed{-\frac{63}{65}}\)