You should use MOSFETs even at 150W because of the increase in efficiency and decrease in losses over BJTs.
Let's say you're making a 12V 150W inverter. So current should be 12.5A. But you have to consider the losses in the inverter. Let's assume the efficiency of the converter is 75%. Also, you have to remember that the battery voltage will be lower than 12V for a large amount of time, depending on the load and capacity of the battery.
Let's assume the battery voltage drops all the way to a 10V minimum (although it should not do so in practice). So, max average current would be (150/10)/0.75 = 20. If we're using push-pull configuration, peak current would be about 1.5*20 = 30A.
So, the MOSFET combination in the inverter should be able to take 30A current. Of course other factors such as gate driving, gate discharge, heat sinking, etc will be important.
So, we can use IRFZ44N as it can handle upto 49A. Power loss in the IRFZ44N would be about 20*20*0.0175 = 7W. You might be happy with that and provide good enough cooling. However, keep in mind that 49A rating is for 25'C. At 100'C it is 35A. So you could parallel one more IRFZ44N and pull the overall loss down to 3.5W plus increased current handling capacity of 35*2 = 70A at 100'C. So, you're completely safe. So, you'll need 4 IRFZ44N's, 2 parallel per push-pull "leg". But if you use IRF3205, you can use 1+1 and still have lower losses. So, you'll have lower losses, better efficiency and you'll need 2 less MOSFETs.
Hope this helps.
Tahmid.
---------- Post added at 12:38 ---------- Previous post was at 12:37 ----------
No. Each branch should be able to handle 1000W. Since when one branch is on, all the current flows through that leg and that leg needs to withstand all the power.