How to plot magnitude and phase response by hand if I have the Transfer Function?
Use the transformation $z = e^{j\omega}$, you will get

$$H(e^{j\omega}) = \frac{1-e^{-j\omega}}{5\left(1+2e^{-j\omega}\right)}$$

Solving this (using $e^{j\omega} = \cos \omega + j \sin \omega$), you should get something like (please double check):

$$H(\omega) = \frac{\left(\cos\omega-1 \right) + j4\sin\omega}{5\left(5+4\cos\omega\right)}$$

Here,

$$\text{Re}\left(H(\omega)\right) = A = \frac{\left(\cos\omega-1 \right)}{5\left(5+4\cos\omega\right)}$$

and, $$\text{Im}\left(H(\omega)\right) = B = \frac{4\sin\omega}{5\left(5+4\cos\omega\right)}$$

Now, the magnitude response will be 

$$\left|H(\omega)\right| = \sqrt{A^2+B^2}$$

and the phase response will be

$$\angle{H(\omega)} = \tan^{-1}\frac{B}{A}$$

You will get both responses as a function of $\omega$, just vary $\omega$, calculate the values and plot the response.