If I have a parallel circuit of 2 resistances, both zero Ohm, the current would split to 2 and each current would be half.
"Zero ohms" is an idealization. If you say you have two zero-ohm resistors in parallel, it just means your model is not accurate enough to determine how the current is split.
If the resistor R1 would be 1 pico Ohm, and the ammeter zero Ohm, there would be no current through the resistor, and all the current would go through the ammeter?
If "zero ohms" means much much less than 1 picoohm, then yes, essentially all the current would go through the ammeter.
But real ammeters have burden resistance that's much much more than a picoohm (more like a few milliohms).
2) Ohms law.
R2 is 10 Ohm, current is 1A.
voltage between point A and B is 0 volt, resistance between A and B is 0 Ohm,
according to Ohms law, I = V / R = 0 / 0 = 0 A.
You have a false conclusion. Zero divided by zero is not zero. It is an undetermined value. Could be zero or could be infinite, depending on the situation.
To analyze this circuit, restate Ohm's law as V = I R. You know the current is 1 A due to the other circuit elements. You know the voltmeter doesn't pass current. Therefore there's 1 A passing from B to A, and because it's a perfect wire, the voltage is zero.
Your estimate is off by several orders of magnitude. Wikipedia gives the resistivity of air as being around \$10^{16}\ \Omega \cdot m\$. I'd guess an actual resistance between two points would be at least on the order of teraohms. Assuming \$1\ T\Omega\$, that gives a current of 5 picoamps, which is far too small to measure easily. As pointed out in an answer to another EE.SE question, the material the battery is made of is probably a better conductor than air.
To actually figure out what's going on in extreme situations, you need a more detailed model of the materials involved. How many electrons and/or ions are available for conduction? An ideal dielectric (insulator) has no free electrons, but a real dielectric might. What's the strength of the electric field? If you have a 40 kilovolt voltage source, you can rip apart air molecules, creating lots of free electrons! A less extreme example would be a vacuum tube, which "conducts" through empty space \$(R = \infty)\$ using electrons liberated from a piece of metal.
Ohm's law is an approximation that works for many materials at low voltages, frequencies, and temperatures. But it is far from a complete description of electrodynamics and physical chemistry, and should not be treated as such.
To answer your question more directly, regardless of whether a tiny current flows through the air, there can definitely be a voltage between the terminals. Voltage is another way of describing the electric field. Wherever there is an electric field, there is a voltage difference, even in a vacuum with no matter at all! HyperPhysics shows what this looks like.
Specifically, the gradient of the voltage field gives you the magnitude and direction of the electric field:
$$\vec E = -\nabla V$$
I don't know whether a tiny current actually flows through the air, but hopefully now you have a better appreciation for the physics of the situation. :-)
Best Answer
If the equivalent resistance is zero then there's also no voltage across it, per Ohm's Law. Then the current though the resistor is 0 V/ 5 Ω = 0 A. The current through the wire can't be calculated this way since 0 V/ 0 Ω is undefined.
Then the current will depend on the source's internal resistance. If that's 1 µΩ for instance the current will be 20 V/ 0.000001 Ω = 20 MA.
If the source has zero resistance current will be infinite.
Either way, applying this to the current divider formula gives for the resistor path:
\$ I_R = I_{tot} \dfrac{R_{tot}}{R} = I_{tot} \dfrac{0 \Omega}{5 \Omega} = 0 A \$
For the wire we get again
\$ I_W = I_{tot} \dfrac{R_{tot}}{R_W} = I_{tot} \dfrac{0 \Omega}{0 \Omega} = undefined \$
And we'll have to look at the external conditions to see how high the current is.
Edit
It is, and mathematicians aren't happy with it either, but there's no other way. Any real thing you try leads to contradictions. Even the 0 volt that I claimed. (I know, I lied, but that was because otherwise I'd get dizzy. Ah, what the heck, let's go for dizziness.)
The voltage across the wire||resistor is
\$ V = \dfrac{R_{PSU}}{R || R_W + R_{PSU}} 20 V = \dfrac{0 \Omega}{0 \Omega + 0 \Omega} 20 V = undefined \$
I can't help it. But let's for the sake of argument say it's 10 V. 0 V didn't get us anywhere, and it has to be between 0 V and 20 V. Then the current through the resistor is 10 V/ 5 Ω = 2 A. The current through the wire is 10 V/ 0 Ω = \$\infty\$ A.
If we apply KCL:
\$ I_{tot} = I_R + I_W \$
That's
\$ \infty A = 2 A + \infty A \$
So far so good. But if we want t0 find \$I_R\$ from this we'll see that we can't! Despite the fact that we know it's 2 A. Let's try:
\$ I_R = \infty A - \infty A = undefined\$
Yeah, right, I always say undefined. Why would it be, if we know the result? OK, you're asking for it. So suppose
\$ \infty - \infty = 2 \$
Now we know that \$\infty\$ + \$\infty\$ = \$\infty\$, so
\$ (\infty + \infty) - \infty = 2 \$
or
\$ \infty + (\infty - \infty) = 2 \$
The value between the brackets is 2, that was our assumption. Then
\$ \infty + 2 = 2 \$
Subtract 2 from both sides, and
\$ \infty = 0 \$
which obviously isn't true. So our assumption was false. Now you can try with any number instead of 2 you'll always end up with a contradiction. So that's how we end up with undefined stuff and a dizzy head.