Let $\theta$ be angle between $\mathbf{a}$ and $\mathbf{b}$ then $\theta=60^{\circ}$ (given)
$$
\begin{aligned}
\text{Since,} \quad |\mathbf{a}+\mathbf{b}|^{2} &=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2 \mathbf{a} \cdot \mathbf{b} \\
&=4+4+\left(2 \times 2 \times 2 \times \cos 60^{\circ}\right) \\
&=8+8 \cos 60^{\circ}=8+4=12
\end{aligned}
$$
$$
\Rightarrow \quad|\mathbf{a}+\mathbf{b}|=\sqrt{12}=2 \sqrt{3}
$$
Now, $\mathbf{a} \cdot(\mathbf{a}+\mathbf{b})=|\mathbf{a}||\mathbf{a}+\mathbf{b}| \cos x$ where $x$ is angle between $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}$.
$\Rightarrow \quad \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}=4 \sqrt{3} \cos x$
$\Rightarrow 4+2 \times 2 \cos 60^{\circ}=4 \sqrt{3} \cos x$
$\Rightarrow \quad 6=4 \sqrt{3} \cos x$
$\Rightarrow \quad \cos x-\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow x=30^{\circ}$