Chọn phương án A.

Giả sử \(z=a+bi\). Khi đó

• \(\sqrt{a^2+b^2}=2\Leftrightarrow a^2+b^2=4\Leftrightarrow a^2=4-b^2\)
• \(z^2+1=\left(a^2-b^2+1\right)+2abi\)

Theo đề bài ta có $$\begin{eqnarray*}
&\left|z^2+1\right|&=4\\
\Leftrightarrow&\sqrt{\left(a^2-b^2+1\right)^2+(2ab)^2}&=4\\
\Leftrightarrow&\left(a^2-b^2+1\right)^2+(2ab)^2&=16\\
\Leftrightarrow&\left(a^2-b^2+1\right)^2+4a^2b^2&=16\\
\Leftrightarrow&\left(4-b^2-b^2+1\right)^2+4\left(4-b^2\right)b^2&=16\\
\Leftrightarrow&\left(5-2b^2\right)^2+16b^2-4b^4&=16\\
\Leftrightarrow&25-20b^2+4b^4+16b^2-4b^4&=16\\
\Leftrightarrow&b^2&=\dfrac{9}{4}\\
\Leftrightarrow&\begin{cases}
b=\pm\dfrac{3}{2}\\
a=\pm\dfrac{\sqrt{7}}{2}.
\end{cases}
\end{eqnarray*}$$
Mặt khác:

• \(z+\overline{z}=(a+bi)+(a-bi)=2a\)
• \(z-\overline{z}=(a+bi)-(a-bi)=2bi\)

Do đó $$\begin{aligned}
\left|z+\overline{z}\right|+\left|z-\overline{z}\right|&=2\left(|a|+|b|\right)\\
&=2\left(\dfrac{\sqrt{7}}{2}+\dfrac{3}{2}\right)\\
&=3+\sqrt{7}.
\end{aligned}$$