Answer
$$\cos(s+t)=\frac{2-2\sqrt{10}}{9}$$
$$\cos(s-t)=\frac{-2-2\sqrt{10}}{9}$$
Work Step by Step
To find $\cos(s+t)$ and $\cos(s-t)$, cosine sum and difference identities need to be applied.
Yet, both cosine sum and difference identities rely on the fact that you already know $\cos s$, $\sin s$, $\cos t$ and $\sin t$.
Therefore, the first job in this type of exercise is to find out all 4 values: $\cos s$, $\sin s$, $\cos t$ and $\sin t$.
1) Find $\cos s$, $\sin s$, $\cos t$ and $\sin t$.
As $\sin s$ and $\sin t$ are known, we would use Pythagorean Identities for $\sin$ and $\cos$ to find the rest:
$$\cos^2 s=1-\sin^2 s=1-\Big(\frac{2}{3}\Big)^2=1-\frac{4}{9}=\frac{5}{9}$$
$$\cos s=\pm\frac{\sqrt{5}}{3}$$
$$\cos^2 t=1-\sin^2 t=1-\Big(-\frac{1}{3}\Big)^2=1-\frac{1}{9}=\frac{8}{9}$$
$$\cos t=\pm\frac{\sqrt8}{3}=\pm\frac{2\sqrt2}{3}$$
Now about the signs of $\cos s$ and $\cos t$:
As $s$ is in quadrant II, $\cos s\lt0$. At the same time, as $t$ is in quadrant IV, $\cos t\gt0$, so $$\cos s=-\frac{\sqrt{5}}{3}\hspace{2cm}\cos t=\frac{2\sqrt2}{3}$$
2) Now we can apply cosine sum and difference identities:
$$\cos(s+t)=\cos s\cos t-\sin s\sin t=\Big(-\frac{\sqrt5}{3}\Big)\Big(\frac{2\sqrt2}{3}\Big)-\frac{2}{3}\Big(-\frac{1}{3}\Big)$$
$$\cos(s+t)=-\frac{2\sqrt{10}}{9}+\frac{2}{9}=\frac{2-2\sqrt{10}}{9}$$
$$\cos(s-t)=\cos s\cos t+\sin s\sin t=\Big(-\frac{\sqrt5}{3}\Big)\Big(\frac{2\sqrt2}{3}\Big)+\frac{2}{3}\Big(-\frac{1}{3}\Big)$$
$$\cos(s-t)=-\frac{2\sqrt{10}}{9}-\frac{2}{9}=\frac{-2-2\sqrt{10}}{9}$$
