Please help me, I'm having a hard time understanding this question and doing it.

Hi Sarcasticcarma!

$cos(\theta)=-\dfrac{\sqrt{2}}{6} \text{ where } \pi \le \theta \le \dfrac{3\pi}{2}$

$tan(\beta)=\dfrac{5}{12} \text{ where } 0 \le \beta \le \dfrac{\pi}{2}$

$\text{What is the exact value of } sin(\theta+\beta)?$

For this question, we need to know the expansion of $sin(A+B)$ in general.

So remember: $sin(A+B)=sin(A)cos(B)+sin(B)cos(A)$

So: $sin(\theta+\beta)=sin(\theta)cos(\beta)+sin(\beta)cos(\theta)$                                                                                       (1)

To solve this question, we just need to find: $sin(\theta),cos(\beta),sin(\beta),cos(\theta)$  (But we already have $cos(\theta)$).

If $cos(\theta)=-\dfrac{\sqrt{2}}{6}$, then we have two ways to get $sin(\theta)$.

First way: Recall: $cos^2(\theta)+sin^2(\theta)=1 \iff sin(\theta)=\pm \sqrt{1-cos^2(\theta)}$

So: $sin(\theta)=\pm \sqrt{1-\dfrac{{2}}{36}}=\pm\sqrt{\dfrac{17}{18}} \\ $ But shall we take the positive or negative?
We must look at the given interval of $\theta$ which is between $\pi \text{ and } \dfrac{3\pi}{2}$

This is the third quadrant, so $sin(\theta)$ is negative. (Determined by the "CAST" rule or drawing sine graph.)

Thus, $sin(\theta)=-\sqrt{\dfrac{17}{18}}$

The second way to get $sin(\theta)$ is much faster and in fact, we will use it to get $cos(\beta),sin(\beta)$, as follows:

Given: $tan(\beta)=\dfrac{5}{12}$

Then, draw a right angle triangle, and choose any angle to be $\beta$

Then, we know $tan(\beta)=\dfrac{5}{12}$, so the opposite side to the angle you chose is 5 and the adjacent must be 12; and by pythagoras theorem, the hypotenuse is 13. (Draw it!)

Now what is $sin(\beta)?,cos(\beta)?$ This is easily done via the triangle we already drew!
So: 

$sin(\beta)=\dfrac{5}{13}\\ \\ \\ cos(\beta)=\dfrac{12}{13}$

And since we are in the first quadrant, all of them will be positive.

So finally by (1):

$sin(\theta+\beta)=sin(\theta)cos(\beta)+sin(\beta)cos(\theta) \\ sin(\theta+\beta)=-\sqrt{\dfrac{17}{18}}*\dfrac{12}{13}+\dfrac{5}{13}*(-\dfrac{\sqrt{2}}{6})$

You can simplify it :).

I hope this helps!