Electronic – 2N7000 Mosfet destroyed during soldering

You want the negative end of the 5V supply (let's call it ground) connected to your MOSFET source, not the drain.
To turn on, you need to apply a voltage between gate and ground, 3V should work okay for most small MOSFETS. You would need to connect the negative side of your 3V to ground (i.e both 3V and 5V negatives tied together) and positive to the MOSFET gate.

Alternatively you can just apply 5V to the gate (the same 5V used to drive the motor)

Also, if you are just touching the voltage to the gate (i.e. not driving with uC or something) then you will need a pulldown resistor between gate and ground to make sure it turns off when power is removed. Something like 10k will do (if you don't have that value, try anything between say, 1k and 100k)

As Faken mentions, a reverse biased diode across the motor is needed to prevent the voltage spike on switch off destroying the transistor. Connect e.g. a 1N4002, cathode to V+, anode to MOSFET drain.

For clarity, here is an example circuit:

Your motor is driven from 5V, so you just put your supply where the 9V supply is. To drive you apply a voltage (above MOSFET turn on) to the gate (FROM MCU) Check your datasheet for the turn on voltage, but 3V or 5V should probably work fine (note with part number shown more than 3V will be needed for reasonable turn on)

Under electrical characteristics in the datasheet, you are looking for a graph like the one shown below. Along the bottom is the drain-source voltage, along the vertical axis is the drain-source current, and each line is a different gate voltage.

Your drain source voltage is 5V. We can see if we apply 3V to the gate we will only get around 30mA, as the MOSFET is not turned on fully. Raising the gate voltage to 4V we will get around 400mA, which should be enough to drive a small motor. Note that the maximum drain source current is only 200mA for this part, so you need to make sure your motors current rating is less than this. If you need more than this then the part shown is no good.

If you give details on the MOSFET and motor used (part numbers, datasheets) more detail can be given.

No, also for P-channel MOSFETs it's charging them. The confusion probably stems from the voltage you see at the lower side, which goes to ground (or near it). But that voltage isn't important, a capacitor's charge is determined by the voltage across it:

\$ Q = \Delta V \cdot C \$

So decreasing the gate voltage increases the gate-source voltage difference, which increases the charge of the capacitor.

When you switch off T1, there's current flowing from +12 V through R2 to the gate to discharge it's capacitance.

edit re the update of your question dd. 2012-07-09(*)
Turning off means that you discharge the gate to +5 V, and this happens by current through R2 and D3. So you bypass R? but R2 is still the limiting factor. A solution would be to swap R2 and T1, so that there's more current/less resistance to discharge the gate than to charge it.

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_{(*) I'm using the ISO 8601 standard date format here. We have user from all over the world and for some 9/07 means 9 July, for others it's 7 September. ISO 8601 is unambiguous.}