What's the rationale behind this choice of resistor for a BJT common collector amplifer?
By the problem description, the amplifier is to source a 35mA sinusoidal current through the load.

Note that during the positive half of the cycle, the output current is from left to right through the capacitor but, during the negative half, the current is from right to left.

Note also that the BJT current is (assuming the transistor is active) always out of the emitter.

Thus, during the negative half cycle, the current through the emitter resistor is the sum of the load current and BJT emitter current.

If follows that the current 'down' through the emitter resistor at the negative peak is at least \$35ma\$.  This would be the case if the BJT were just 'turning off' at the negative peak.

Since it is already given that the emitter voltage should be \$7.22V\$ at the negative peak, and we know the resistor current is at least \$35mA\$ then, Ohm's Law gives

$$R_E \le \frac{7.22V}{35mA}$$

In other words, we have established an upper bound on the value of \$R_E\$.




  We want maximum voltage swing, so we say the emitter DC voltage is
  approximately 7.5V, half of our positive rail...


For sinusoidal operation and assuming the capacitor voltage is constant (the capacitor is an AC short circuit) and equal to the DC emitter voltage \$V_E\$, we have:

$$v_{O+} = V_+ - V_E $$

$$v_{O-} = -V_E \frac{R_L}{R_E + R_L} $$

Typically, one wants these to be equal in magnitude, i.e., we want the output voltage swing to be symmetric.

Setting \$|v_{O+}| =  |v_{O-}|\$ yields the following condition for symmetric load voltage swing:

$$V_E = V_+ \frac{R_E + R_L}{R_E + 2R_L} $$