Insert a 100 ohm or so resistor in series with the LED, to limit the current if you get the transistor fully turned on.
Disconnect the audio signal for now.
Instead of your BE and BC resistors, use a pair of resistors, one from B to +5, and the other from B to ground. The ratio should bias the base so that the LED is just dark. That should be about 10k to +5 and 1k to ground -- but adjust until the LED is slightly off.
4 Connect the audio to the base via a capacitor -- this is actually the crucial ingredient. A range of values will probably work -- say 10uF. That will be polarized -- the + end should go to the base.
That should be much more sensitive than what you've got. Your main issue is that the audio is likely at zero volts, and so you need about 0.5V to 0.7V peaks (on the positive side) to get the transistor to turn on. Using biasing resistors and AC coupling with the cap means you need much smaller +ve peaks to get some LED lighting.
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.
Best Answer
The major limit to BJT switching time is related to the charge carriers and specifically how long it takes to move carriers into the base, and how long it takes to get them out.
The datasheet will include a few parameters that will give you the theoretical maximum switching frequency*. They are
Using the datasheet (these parameters are usually listed), you can figure out how fast a transistor can switch between the two states.
$$ f_{max} = \frac{1}{t_d + t_r + t_f + t_s} $$
* This is what transistor can theoretically do, but there are tricks that can be done to improve the switching speed. Also, if you are switching a square wave, then in order to maintain a nice square waveform, the actual switching frequency will be much less.