1) Yes
2) Correct **
3) 0,045Ohmx1,5A=0,0675V. 0,0675V*1,5A=0,1W, so it's OK
4) That is maximal allowable voltage between D and S to still operate as transistor. When the transistor is off (Vgs=0V), the drain "is seeing" the full supply voltage, in this case 12V*. Once you apply more than 20V between D and S, manufacturer guarantees nothing and anything can happen. From nothing to black hole forming and swallowing universe.
Very likely, it will survive few (tens) volts of overloading, then it starts to behave either as conductor or insulator, often in non-reversible way.
* 20V MOSFET could too low (with respect to Vds) for automotive environment, where voltage spikes on 12V rail could be way higher. Especially when switching motors or other inductive loads.
------
** EDIT: it is 0,045Ohm typical, 0,057Ohms maximal. You always have to deal with worst-case scenarios, so take 0,057Ohm figure. The calculation prom point 3 then gives you 0,13W, which is still OK.