Let:
\[ z + 2 = \cos \theta + i \sin \theta \implies \frac{1}{z + 2} = \cos \theta - i \sin \theta. \]
Now:
\[ \frac{z + 1}{z + 2} = 1 - \frac{1}{z + 2} = 1 - (\cos \theta - i \sin \theta). \]
Simplify:
\[ \frac{z + 1}{z + 2} = (1 - \cos \theta) + i \sin \theta. \]
The imaginary part is:
\(\ Im \left( \frac{z + 1}{z + 2} \right) = \sin \theta = \frac{1}{5}.\)
Using \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left( \frac{1}{5} \right)^2 = 1 - \frac{1}{25} = \frac{24}{25}. \]
\[ \cos \theta = \pm \sqrt{\frac{24}{25}} = \pm \frac{2 \sqrt{6}}{5}. \]
Now, the real part of \(z + 2\) is:
\(\ Re(z + 2) = \cos \theta.\)
The magnitude of \(\Re(z + 2)\) is:
\[ |\ Re(z + 2)| = \frac{2 \sqrt{6}}{5}. \]
Final Answer: \(\frac{2 \sqrt{6}}{5}\).