Step 1 :Given that $\cos s=-\frac{4}{5}$ and $\sin t=\frac{2}{5}$, we can use the Pythagorean identity to find $\sin s$ and $\cos t$.
Step 2 :Since $s$ is in quadrant II, $\sin s$ is positive. So, $\sin s=\sqrt{1-\cos^2 s}=\sqrt{1-\left(-\frac{4}{5}\right)^2}=\frac{3}{5}$.
Step 3 :Similarly, since $t$ is in quadrant II, $\cos t$ is negative. So, $\cos t=-\sqrt{1-\sin^2 t}=-\sqrt{1-\left(\frac{2}{5}\right)^2}=-\frac{\sqrt{21}}{5}$.
Step 4 :We can now use the cosine of a sum and cosine of a difference identities to find $\cos (s+t)$ and $\cos (s-t)$.
Step 5 :Using the identity $\cos (s+t)=\cos s \cos t - \sin s \sin t$, we find $\cos (s+t)=\left(-\frac{4}{5}\right)\left(-\frac{\sqrt{21}}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)=\frac{4\sqrt{21}}{25}-\frac{6}{25}=\frac{4\sqrt{21}-6}{25}$.
Step 6 :Using the identity $\cos (s-t)=\cos s \cos t + \sin s \sin t$, we find $\cos (s-t)=\left(-\frac{4}{5}\right)\left(-\frac{\sqrt{21}}{5}\right)+\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)=\frac{4\sqrt{21}}{25}+\frac{6}{25}=\frac{4\sqrt{21}+6}{25}$.
Step 7 :So, $\cos (s+t)=\boxed{\frac{4\sqrt{21}-6}{25}}$ and $\cos (s-t)=\boxed{\frac{4\sqrt{21}+6}{25}}$.