Power losses of a switch
Those MOSFET formulas don't take into account any switching; only the time in between - I'm pretty sure you knew that, but anyway.

When the transistor is on, it has 2v across it and 25A going through it. 


  P = I V = 25 x 2 = 50W.


But duty cycle is 50%, so during the on period it dissipates 50W, but overall, 25W. Actually, it's a but less than that, since there is the turn off / on time of 100ns.

The total cycle time = 1 / 100kHz = 10us, so with switch time being 100ns, that's 1% each way. So, your transistor is only fully on for 49% of the time. Substitute as above, and you get 24.5W for the fully on portion.

When off, no current flows, therefore no power is dissipated. So that's zero watts.

Finally, the rise and fall time. This could get very tricky depending how far you want to go; your exercise gives little hint as far as I gather what's expected; only the "inductive" load for which AFAIK detail are missing; that, and that the model of the transistor is highly simplified, so they're trying to get you to focus on something other than that. So, guessing, but maybe they want the following.

So, rise and fall: it will dissipate a lot if power during this transition period; consider when half the voltage is across the transistor, and the other half across a resistive load. The current through both would be 12.5A. At 300v, each will dissipate 12.5 x 300 = 3750W (at that instant). That would be the peak power dissipating in the transistor (impedance matching) so that gives you an upper bound for switching losses. 2% of the time, no more than (2% x 3750W) = 75W would be lost. So the total losses would no more than 99.5W. But as I said, that assumes a resistive load.