Your results is partly correct. The angle $\theta$ is correct. I'm finding a different at the second part.
Lets write $F_1 = 800[N]$ and $F_2=500N$.
Then the resultant R will have:
$$R_x = F_1 + F_2 \cos(\phi) \qquad R_y=F_2 \sin(\phi)$$
Then $$|R|= \sqrt{(F_1 + F_2 \cos(\phi))^2 + (F_2 \sin(\phi))^2 }$$
Solving for this you get
$$a= 97.09[deg]$$
Therefore:
- $R_x = 495.25 [N]$
- $R_y = -868.75 [N]$ (Negative because its pointing to the left)
and the angle between the resultant and the horizontal +x (assuming +x is to the right) is:
I'll put up the answer when you show what you've tried
$$\beta = 150.31[deg]$$