$\cos(x)=-\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$. Calculate $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$. If $\cos(x)=  -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$?

Solution:
If $\frac{\pi}{2} < x < \pi$ then right triangle with acute angle $x$ facing west at second quadrant, the hypotenuse must be positive and the only negative is the adjacent. 
$\cos(x)=  -\frac{24}{25}$ then $adjacent=-24$ and $front=7$ so $\sin(x) = \frac{7}{25}$.  
If $\pi < y <\frac{3\pi}{2}$ then right triangle with acute angle $x$ facing west at third quadrant, the front and adjacent are negative. If $\tan(y) = \frac{9}{40}$ then the front is $-9$, the adjacent is $-40$, and the hypotenuse is $41$. So $\sin(y) = -\frac{9}{41}$ and $\cos(y)=-\frac{40}{41}$
The right triangle that we consider is the one with the adjacent side on the $x$-axis right..?

Your method is correct. Indeed, refer to the graphs:
$\hspace{1cm}$
$$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$
Also note:
$$\cos x=-\cos(\pi -x)=-\frac{24}{25} \Rightarrow \cos (\pi -x)=\frac{24}{25}; \quad \sin (\pi -x)=\frac{7}{25}=\sin x.\\
\tan y=\tan(y-\pi)=\frac{9}{40}=\frac{-9}{-40} \Rightarrow \begin{cases}\sin y=-\sin(y-\pi)=-\frac{9}{41} \Rightarrow \sin (y-\pi)=\frac9{41}\\ \cos y=-\cos(y-\pi)=-\frac{40}{41} \Rightarrow \cos (y-\pi)=\frac{40}{41}\end{cases}$$
Alternatively, you can remember the signs of sine, cosine, tangent and cotangent functions in the four quarters: sine (++--); cosine (+--+); tangent and cotangent (+-+-).
