BJT Differential amplifier. Tail Resistance. Input Resistance
How did they get these values?


For Diff amplifier the symmetrica input resistance is equal to: 

\$ R_{IN} = \beta(2 \times r_e) = 100 * 500\Omega = 50 \textrm{k}\Omega\$

Or

\$ R_{IN} = \beta \left(2 \times (r_e + R_E)\right) = 100 * 2.5\textrm{k}\Omega = 250 \textrm{k}\Omega\$

Also, look here: 
BJT Differential Amplifier Common Mode &amp; Differential Mode Gain


  Why R1 is not in the input impedance formulas, although it is in
  series with emitter resistors with respect to ground?


As for your question about R1 influence. Try to find Rin for this circuit: 



I hope that you see that without any big error we can say that \$ Rin = 2R_E\$ because the lower \$R_E\$ resistor is connected in parallel with \$R_1\$, but usually, \$R_1 &gt;&gt; R_E\$ so can ignore \$R_1\$ effect on asymmetrical input resistance (when Q2 base is connected to GND).   


  Are there any drawbacks, namely are we losing something by making R1
  -> infinity?


In short. The main job of a diff amplifier is to amplify only the voltage difference between two inputs. And the no amplification for a common-mode input voltage (the same voltage seen at both inputs). We want\$G_{CM} = 0 V/V\$ and \$CMRR = \infty\$ 
And to accomplish this task we replace \$R_1\$ with an ideal current source. 
And the ideal current source will have \$r_o =  \infty \$ (R1) and this gives us the common-mode gain equal to \$0\$ and \$CMRR = \infty\$.