[tex]\theta[/tex] is in quadrant I, so [tex]\cos\theta>0[/tex].
[tex]x[/tex] is in quadrant II, so [tex]\sin x>0[/tex].
Recall that for any angle [tex]\alpha[/tex],
[tex]\sin^2\alpha+\cos^2\alpha=1[/tex]
Then with the conditions determined above, we get
[tex]\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35[/tex]
and
[tex]\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}[/tex]
Now recall the compound angle formulas:
[tex]\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta[/tex]
[tex]\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta[/tex]
[tex]\sin2\alpha=2\sin\alpha\cos\alpha[/tex]
[tex]\cos2\alpha=\cos^2\alpha-\sin^2\alpha[/tex]
as well as the definition of tangent:
[tex]\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}[/tex]
Then
1. [tex]\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}[/tex]
2. [tex]\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}[/tex]
3. [tex]\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}[/tex]
4. [tex]\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}[/tex]
5. [tex]\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}[/tex]
6. [tex]\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7[/tex]
7. A bit more work required here. Recall the half-angle identities:
[tex]\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2[/tex]
[tex]\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2[/tex]
[tex]\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}[/tex]
Because [tex]x[/tex] is in quadrant II, we know that [tex]\dfrac x2[/tex] is in quadrant I. Specifically, we know [tex]\dfrac\pi2<x<\pi[/tex], so [tex]\dfrac\pi4<\dfrac x2<\dfrac\pi2[/tex]. In this quadrant, we have [tex]\tan\dfrac x2>0[/tex], so
[tex]\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32[/tex]
8. [tex]\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}[/tex]
