(a) $\dfrac{1}{\sqrt{2}}$
Given $\cos(\alpha+\beta)=0$, so $\alpha+\beta=90°$. Thus $\dfrac{\alpha+\beta}{2}=45°$, and $\cos 45°=\dfrac{1}{\sqrt{2}}$.
Since $\cos\theta=0$ gives $\theta=90°$, we get $\alpha+\beta=90°$, so $\dfrac{\alpha+\beta}{2}=45°$. From the standard table, $\cos 45°=\dfrac{1}{\sqrt{2}}$. The key is recognising which angle has cosine equal to zero.