Class A Emitter follower instantaneous power
Why the maximum instantaneous power dissipated by Q1 is \$P=2V_{cc}*I\$ at RL
  open?


\$v_I\$ causes \$v_O\$ to be at a minimum value when \$v_I\$ is at a minimum value. This is because the transistors are basically emitter followers. Under these circumstances, approximately all the supply voltage (\$2\cdot V_{CC}\$) appears across Q1 and so Q1's power dissipation is \$2\cdot V_{CC}\cdot I\$.

It's an approximation because Q2 will never fully turn on to 0 volts and hence, there might be 200 mV less across Q1 and not the full \$2\cdot V_{CC}\$.