Obtaining the magnitude of the frequency response by plugging $e^{jω}$ into the z-domain transform function?
It's  derived using Euler's equation http://en.wikipedia.org/wiki/Euler's_formula. You start with $$H(z) = (\frac{1}{1 - \frac{1}{5}z^{-1}}) (\frac{1}{1 + \frac{1}{2}z^{-1}})$$

Then plug in $e^{-jwT}$ for $z^{-1}$ and you get 

$$H(z) = (\frac{1}{1 - \frac{1}{5}e^{-j\frac{\pi }{2}}}) (\frac{1}{1 + \frac{1}{2}e^{-j\frac{\pi }{2}}})$$

Now according to Euler we simply have $e^{-j\frac{\pi }{2}} = -j$ so the whole thing simplifies to

$$H(z) = (\frac{1}{1 - \frac{1}{5}\cdot (-j)}) (\frac{1}{1 + \frac{1}{2}\cdot (-j)})$$. 

Utilizing $(-j) \cdot j = 1$ we can multiple each part of the fraction with j/j and get 
$$H(z) = (\frac{j}{j - \frac{1}{5}}) (\frac{j}{j + \frac{1}{2}})$$

Then we can take the magnitude as 

$$\left | H(z) \right | = \frac{\left | j^{2} \right |}{\left | j-\frac{1}{5} \right |\cdot \left | j+\frac{1}{2} \right |}$$

This isn't exactly the same formula that you have given but since magnitude j and j^2 are the same, the final result comes out to be the same.