Electronic – Motor power rating and in-rush current

The 180W reasoning is wrong. You don't have the full 12V across the fuse, only a fraction of it. Say the fuse has a 0.1\$\Omega\$ resistance, then 15A will cause a power dissipation of

\$ P = I^2 \cdot R = (15A)^2 \cdot 0.1 \Omega = 22.5W \$.

And that 22.5W in the thin wire is just too much, and the fuse will blow.
Note that the 12V doesn't appear anywhere in this, and you could do the same calculation for 15A at 230V AC. There's a "but", however. Fuses do have a rated voltage. Not for the blowing, as we've seen, but to make sure the remains of the fuse won't cause sparks when at a high voltage. So you can't use a 12V car fuse in a mains appliance, but the other way around should be safe.

Automatic fuses also react on current only. They have a coil which releases a latch if the number of coil-windings exceeds the rated value. Also here it's independent of voltage.

The 12 V 3 A supply sounds appropriate, but the 5 V supply is not. If you want to maximize the holding torque, you have to allow the maximum current thru the windings. This can't be achieved at only 5 V according to the specs you quoted.

Added:

Now that you have supplied a link to the datasheet, we know the specs for sure. Your motor is rated for 1 A, 3.8 Ω DC resistance, 3.8 mH inductance, and 75 V max. I have no idea where you got the 12 V ratings from you mentioned in your question.

Since the maximum current is 1 A and the coil has a resistance of 3.8 Ω, you can only apply 3.8 V to a winding continuously. The 75 V rating is the maximum you must not exceed under any circumstances.

Since the motor coils have significant inductance, it takes time to build up current when a fixed voltage is applied. However, torque is proportional to current. This is why stepper motors are often driven with a roughly current controlled supply. The voltage will be high initially as the current in the inductor builds up, then drop to the sustaining level. If you know you are stepping the motor fast, then you can use a higher fixed voltage keeping in mind the time it takes for the current to build to the maximum level given the coil inductance and resistance. Since these parameters are reasonably well known, you can do this open loop. The usual way nowadays is to have a micro do the calculations and control the coil drive with PWM from a fixed high voltage. The duty cycle is changed such that the coil sees a apparent high voltage right after switching, which then decays down to the sustaining voltage as the current builds up. You can also use the same PWM technique but with some current feedback. This feedback is used to adjust the PWM duty cycle to whatever it takes to maintain the desired current.