You are asking what transistor is good for a low side switch that must be able to handle 1A and 12V. This needs to be driven from a CMOS digital output. You say you have no idea what the current output capability of the digital chip is (4013), but that's silly. Surely this is specified in the datasheet. For sake of example only (I didn't look it up, that's you're job especially since you didn't supply a easy link to the datasheet), let's say we don't want to draw more than 5mA from the digital output when it is high. I just noticed that these chips are powered from 12V, so let's say "high" means at least 10V. Again, this is something you need to look up.
A NPN transistor makes a simple and cheap low side switch. It has to be able to handle 1A. That is getting into the low end of the "power" transistor category, so let's say you can find one with a gain of 20 at 1A. That means it will need 1A/20 = 50mA base current minimum. Since that's more than the digital output can source, you need another transistor in there to provide more gain. This first transistor (since the signal goes thru it before the power transistor) only needs to handle a bit over 50mA and 12V. That's "small signal" territory where much higher gains can be had. There will be many cheap transistors that can be counted on to have a gain of 50 in this case, so the digital output now only needs to source 1mA, which is within spec. Here is one way to arrange all this:
One problem with this is that since you are using such a high voltage for the digital chip, the base current for Q2 has to drop accross a higher voltage and therefore dissipate much more power than if the digital output were a more normal 5V or 3.3V when high. There are ways to deal with that, like a divider right after the digital output, but I'm showing the conceptually simple case. For the purpose of computing the dissipation in R1, you have to assume the digital output goes all the way to it's supply, which is 12V. That 12V minus the two B-E drops will be about 10.6V. (10.6V)^2 / 160Ω = 700mW. That's quite a lot and means you have to use at least a 1W resistor.
As I said, there are ways to mitigate the wasted power in driving a NPN low side switch, but this application is really crying out for a FET. Now the 12V digital signal is actually a advantage. There are many many N channel MOSFETs that can handle 12V and turn on to a small fraction of a Ohm with 12V gate drive. This simplifies the drive circuitry, and is what I recommend:
Let's do a quick sanity check on the power dissipation of Q1. Let's say the FET goes down to 100mΩ. (1A)^2 * 100mΩ = 100mW. That's something even a SOT-23 package can handle, and 100mΩ is quite high for a FET of this voltage and current rating.
As for reversing direction, that's just how the coils are sequenced. If you produced the coil signals with a microcontroller as we've been telling you to do all along, this would be a trivial thing to do in firmware.
To determine the current you need, you first have to know the torque you need. That is a function of the external system that only you know. Once you know the torque you look in the table and see what current is required to generate that torque. If you can tolerate a lower torque for holding versus moving, then you can have a lower hold current than run current.
Keep in mind that motion inevitably eats up some torque due to friction, which actually helps for the holding case. The back EMF of the motor will also reduce the apparent voltage applied to the windings as a function of speed. Stepper motors are rather inefficient, so their back EMF is usually not that big. You can measure it yourself by looking at the open circuit voltage when mechanically spinning the motor at a known speed. That's how much less voltage the motor "sees" at that speed than what you apply.
Best Answer
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.