Assuming that enough current flows through both diodes so they regulate, and assuming they're perfect, then the voltage across them is their Vz.
Thus current in the 500R resistor is (100-10)/500 = 180mA
Current in the 300R resistor is (10-6)/300 = 13.3mA
Current in Z1 is (180-13.3) = 166.6mA and power is 1.66W
So, your calculation is correct. Z1 dissipates more than what it is rated for. Note that as analogsystemsrf says, this does not necessarily mean it will blow. This depends how the power rating is calculated. What determines lifetime is maximum chip temperature and thermal cycling / dilatation of materials... and that depends on how the heat is removed from your part. So the same part (for example a TO220 package) can have wildly different power ratings depending how it is mounted (free air, board, heat sink...) or if a fan blows on it or not.
If we assume the power rating is correct, then the diode will overheat. It can melt its solder joints and fall off the board, or simply blow. Leaded zeners mainly fail shorted, at least the modern 'nailhead' types that don't have wire bonds. A SOT-23 type zener does have bondwires, so can blow open after it shorts if enough current is available.
Now, to answer the second question, even with RL=0 ohms, we have 33mA through the 300R resistor, 147mA through Z2, and it still dissipates too much. There's probably a typo in the value of the 500R resistor.... the resistor dissipates 16W which is no good. It should be 5000 ohms.