Determine $\sin(x + y)$ and $\cos(x - y)$, if you know $\cos(x) = \frac{5}{7}, \quad x\in(0,\pi/2), \quad \sin(y) = \frac{1}{5},\quad y\in(\pi/2,\pi)$ I need advice on this example.
Determine $\sin (x + y)$ and $\cos (x - y)$, if you know that
$\cos(x) = \frac{5}{7}, \quad x\in(0,\pi/2), \quad \sin(y) = \frac{1}{5}, \quad y\in(\pi/2,\pi).$
I used this formula to express $\sin (x)$ and $\cos (y)$
$$\sin^2(x)+\cos^2(x)=1$$
I recieved:
$$\sin(x)=\sqrt{24}/7\quad\text{and}\quad\cos(y)=\sqrt{24}/5.$$
However, if I now substitute into these formulas I get a different result than I have in the assignment
$$
\begin{align*}
\sin(x+y)&=\sin(x)\cdot \cos(y)+\cos(x)\cdot \sin(y)\\
\cos(x-y)&=\cos(x)\cdot \cos(y)+\sin(x)\cdot \sin(y)
\end{align*}
$$
The correct result should be
$$\cos(x-y) = -8\sqrt{6}/35\quad\text{and}\quad\sin(x+y) = -19/35.$$

You have a sign error.
$y$ is in the second quadrant, hence $\cos(y) = -\frac{\sqrt{24}}{5}$.
\begin{align}
\sin(x+y) =  -\frac{\sqrt{24}}{7} \cdot \frac{\sqrt{24}}{5}  + \frac57 \cdot  \frac15 = -\frac{19}{35}
\end{align}
