$\left|\frac{1+i \sqrt{3}}{\left(1+\frac{1}{i+1}\right)^{2}}\right|=\left|\frac{(1+i \sqrt{3})(i+1)^{2}}{(i+2)^{2}}\right|\left(\because i^{2}=-1\right)$
$=\left|\frac{(1+i \sqrt{3})\left(i^{2}+1+2 i\right)}{\left(i^{2}+4+4 i\right)}\right|$
$=\left|\frac{(1+i \sqrt{3}) 2 i}{(4 i+3)}\right|$
$=\left|\frac{-2 \sqrt{3}+2 i}{3+4 i}\right|$
$=2\left|\frac{(-\sqrt{3}+i)(3-4 i)}{(3+4 i)(3-4 i)}\right|$
$=2\left|\frac{-3 \sqrt{3}+3 i+4 \sqrt{3} i+4}{25}\right|$
$=\frac{2}{25}|(4-3 \sqrt{3})+(3+4 \sqrt{3}) i|$
$=\frac{2}{25} \sqrt{(4-3 \sqrt{3})^{2}+(3+4 \sqrt{3})^{2}}$
$=\frac{2}{25} \sqrt{16+27-24 \sqrt{3}+9+48+24 \sqrt{3}}$
$=\frac{2}{25} \sqrt{100}=\frac{2 \times 10}{25}=\frac{4}{5}$