$$
\begin{aligned}
& \text { } \because \quad \vec{a}, \vec{b} \text { are unit vectors. So, }|\vec{a}|=1 \&|\vec{b}|=1 \\
& \text { Also, }|\vec{a}-\vec{b}|^2=1 \Rightarrow|\vec{a}-\vec{b}|^2=1 \\
& \Rightarrow|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b}=1 \\
& \Rightarrow 1+1-2 \vec{a} \cdot \vec{b}=1 \Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \\
& |\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1+1+2 \times \frac{1}{2}=3 \\
& \Rightarrow|\vec{a}+\vec{b}|=\sqrt{3} \\
& \because \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \\
& \Rightarrow 2 \cos ^2 \frac{\theta}{2}-1=\frac{1}{2} \Rightarrow 2 \cos ^2 \frac{\theta}{2}=\frac{3}{2} \Rightarrow \cos ^2 \frac{\theta}{2}=\frac{3}{4} \\
& \Rightarrow \cos \frac{\theta}{2}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Now, $2|\vec{a}+\vec{b}| \cos \frac{\theta}{2}=2 \times \sqrt{3} \times \frac{\sqrt{3}}{2}=3$