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 \(\boxed{-\frac{63}{65}}\)