Since [tex]\alpha[/tex] lies in quadrant II and [tex]\beta[/tex] lies in quadrant IV, we expect [tex]\sin\alpha>0[/tex], [tex]\cos\alpha<0[/tex], and [tex]\sin\beta<0[/tex].
Recall the Pythagorean identities,
[tex]\sin^2x+\cos^2x=1\iff1+\cot^2x=\csc^2x\iff\tan^2x+1=\sec^2x[/tex]
It follows that
[tex]\sec\alpha=\dfrac1{\cos\alpha}=-\sqrt{\tan^2\alpha+1}=-\dfrac{13}5\implies\cos\alpha=-\dfrac5{13}[/tex]
[tex]\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{12}{13}[/tex]
[tex]\sin\beta=-\sqrt{1-\cos^2\beta}=-\dfrac45[/tex]
Recall the angle sum identity for sine:
[tex]\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha[/tex]
So we have
[tex]\sin(\alpha+\beta)=\dfrac{12}{13}\dfrac35+\left(-\dfrac45\right)\left(-\dfrac5{13}\right)=\boxed{\dfrac{56}{65}}[/tex]
