Let $\cos \alpha+\cos \beta=\frac{3}{2}$

$\Rightarrow 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{2}$ ….. $(i)$

and $\sin \alpha+\sin \beta=\frac{1}{2}$

$\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{1}{2}$ ….. $(ii)$

On dividing $(ii)$ by $( i)$, we get

$\tan \left(\frac{\alpha+\beta}{2}\right)=\frac{1}{3}$

Given $: \theta=\frac{\alpha+\beta}{2} \Rightarrow 2 \theta=\alpha+\beta$

Consider $\sin 2 \theta+\cos 2 \theta=\sin (\alpha+\beta)+\cos$
$(\alpha+\beta)$

$=\frac{\frac{2}{3}}{1+\frac{1}{9}}+\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{6}{10}+\frac{8}{10}=\frac{7}{5}$