Given: $\sin \beta=2 \sin \alpha...(i)$
and $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha...(ii)$
Now, $\sin ^2 \alpha+\cos ^2 \beta=(2 \sin \alpha)^2+\left(\frac{2}{3} \cos \alpha\right)^2$
$\Rightarrow 1=4\left(\sin ^2 \alpha\right)+\frac{4}{9}\left(1-\sin ^2 \alpha\right)$
$\Rightarrow 1=4 \times \frac{8}{9} \sin ^2 \alpha+\frac{4}{9} \Rightarrow \sin ^2 \alpha=\frac{5}{32}$
Now, $\sec (\alpha+\beta)=\frac{1}{\cos (\alpha+\beta)}=\frac{1}{\cos \alpha \cos \beta-\sin \alpha \sin \beta}$ $=\frac{1}{\frac{2}{3} \cos ^2 \alpha-2 \sin ^2 \alpha}$ \{from equations (i) \& (ii)\} $=\frac{1}{\frac{2}{3}-\frac{8}{3} \sin ^2 \alpha}=\frac{1}{\frac{2}{3}-\frac{8}{3} \cdot \frac{5}{32}} \Rightarrow \sec (\alpha+\beta)=4$