Use the cosine of a sum and $\operatorname{cosine}$ of a difference identities to find $\cos (s+t)$ and $\cos (s-t)$.
\[
\cos s=-\frac{4}{5} \text { and } \sin t=\frac{2}{5}, s \text { and } t \text { in quadrant II }
\]
\[
\cos (s+t)=
\]
(Use fractions or pi for any numbers in the expression.)

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 $\cos (s-t)=\boxed{\frac{4\sqrt{21}+6}{25}}$.