Dissipating 1W on a TO-220 without heatsink?
This is simple: do the math. Look at the datasheet. There should be a thermal resistance spec that tells you how many degC difference there will be between the die and ambient air per Watt. Then add that to your worst case ambient temperature and compare to the maximum allowed die temperature.
For most transistors and ICs, a TO-220 case will get hot at 1W, but generally stay within operating range. At 1/2 W I wouldn't worry about it. At 1W I'd check the datasheet and do the calculation but it will probably be OK.
One wrinkle: The datasheet may only tell you die to case thermal resistance. You then have to add the thermal resistance from the case to ambient, which will be much higher. Fortunately that's mostly a function of the TO-220 case, not the transistor, so you should be able to find a generic figure for that. Good datasheets give you both thermal resistance figures.
Added:
I hadn't followed the datasheet link earlier, but now I see that everything you need is well specified in there. The thermal resistance from die to ambient is 62.5 C/W, and the maximum die operating temperature is 175C. You said your ambient temperature is 25C. Adding the rise from there to the die at 1W yields 88C. That's 87C below the maximum operating temperature, so the answer is very clearly YES, your transistor will be fine at 1W in 25C free air.
Answering your second question:
A switching MOSFET will have two types of losses; conduction and switching. Conduction loss is the usual \$I_D^2 \times R_{DS(on)}\$ loss. If you control the MOSFET so that it's on with 50% duty cycle, the conduction loss is 50% of the DC (always-on) loss.
Switching losses include the amount of energy needed to control the gate and losses in the device as it transitions from the on-state to the off-state. When you're turning on a MOSFET, there is an interval where \$I_D\$ starts to flow and the \$V_{DS}\$ voltage is still at its maximum. \$V_{DS}\$ falls as the MOSFET channel saturates. The power consumed during this time is called turn-on loss. Similarly, at turn-off, there's an interval where \$V_{DS}\$ rises before \$I_D\$ starts falling, which (not surprisingly) is called turn-off loss.
You must consider the turn-on and turn-off losses when you're talking about 100kHz operation. Most likely you will see less power than the DC condition, but you won't be saving 50%.
Answering your third question:
MOSFET \$R_{DS(on)}\$ has a positive temperature coefficient - the warmer it gets, the higher the \$R_{DS(on)}\$ gets. If you connect two MOSFETs in parallel with similar characteristics (i.e. the same part number from the same manufacturer), drive them identically, and don't have huge asymmetry in your PCB layout, the MOSFETs will indeed share current quite nicely. Always make sure each MOSFET has an independent resistor in series with each gate (never parallel gates without resistors) as gates tied directly together can weirdly interact with each other - even a few ohms is better than nothing.
Answering your first question:
let's start with the power consumption. Datasheet says 4.7m\$\Omega\$ maximum at 75A, and at 12.5A this will be less, so that's a safe value. Then \$P = I_D^2 \times R_{DS(ON)} = 12.5^2 \times 4.7m\Omega = 735mW\$. Add some extra safety and 1W is a good value.
What a part can dissipate depends on
- the amount of energy generated,
- how easily the energy can be drained to the environment
(The first factor says "energy", and not "power", because it's energy that causes temperature rises. But in our calculations we assume steady state, and can divide everything by time so that we can work with power instead of energy.)
We know the power, that's 1W. How easily the energy can be drained is expressed in thermal resistance (in K/W). This thermal resistance is the sum of a few different thermal resistances which you normally (should) find in the datasheet: there's the junction-to-case resistance and the case-to-ambient resistance. The former is very low, because the heat transfer is through conduction, while the latter is a much higher value because here the heat transfer is through convection. Like Olin says the latter is a property of the case type (TO-220), so maybe we won't find it in the datasheet. But we're in luck, the datasheet gives us the total thermal resistance, junction-to-ambient: 62.5 K/W. That means that at a 1W dissipation the junction temperature will be 62.5 K (or °C) higher than the environment. If the temperature in the enclosure is 25°C (that's rather low!), then the junction temperature will be 87.5°C. That's much less than the 125°C which is often assumed as a maximum temperature for silicon, so we're safe. The case temperature will be almost the same as the junction, so the MOSFET will be HOT, too hot to touch.
Note: this web page lists the case-to-ambient thermal resistance for different packages.