Can you explain this differential amplifier exercise to me?
VE=-0.7 Volt thus I=(-0.7+5)/4.3=1 mA so re= 50 ohm.Why is re=50 Ohm
  here?


For a single bipolar transistor, \$r_e=\frac{V_T}{I_C}\$, where \$V_T=\frac{kT}{q}=26\rm{mV}\$ at room temperature, and \$I_C\$ is the collector current in Amps. The tail current is \$\frac{4.3\rm{V}}{4.3\rm{k\Omega}}=1\rm{mA}\$. Half of this flows into each transistor, so for each transistor: \$r_e=\frac{26\rm{mV}}{0.5\rm{mA}}=52\Omega\$. For easier calculations, the author rounded to \$50\Omega\$.


  a) Ad=V0/Vd=α*2/2re .Why is Re ommited from this formula? Isnt it
  supposed to be Ad=α*2/(2re+2Re)?


If \$v_{B1}\$ rises the same amount that \$v_{B2}\$ falls, then the voltage at the common node connected to the emitters doesn't move (assuming, of course, linearity -- i.e., small signals). Thus, \$R_e\$ doesn't matter, because the voltage across it remains fixed. You can basically treat the common node as a "ground" of sorts, and imagine Q2 operating as a common-emitter amplifier whose gain is simply \$\frac{v_o}{v_{B2}}=-\frac{2\rm{k\Omega}}{r_e}=-\frac{2\rm{k\Omega}}{50}=-40\frac{\rm{V}}{\rm{V}}\$ (the normal equation for a common-emitter amplifier). Since the signal being applied to its base is \$v_{B2}=-\frac{v_{id}}{2}\$, then the actual gain \$\frac{v_o}{v_{id}}=20\frac{\rm{V}}{\rm{V}}\$.