What is the intuitive explanation for high voltage and low current? By the formula $$P=VI$$ we can say that the voltage can be increased and the current decreased by keeping the power constant. It is mathematically and experimentally right. But is there a way to intuitively explain how this can happen?
Considering that voltage is a measure of tendency of a terminal to send or receive electrons, how can we have a high tendency to do so (high voltage), but low rate of passage of electrons (low current)?
 A: Yes. There are lots of cases where there are large forces but little movement.
For example, there is a large force pulling me down into the ground, but I fail to move at all in that direction, because there is a floor in the way and I cannot pass through the floor. Similarly it is very common to have two conductors at very different voltages next to each other, but there is no conductive pathway between them, so no current flows.
When only a little current is flowing, it is generally because there is something pushing back. For example I might jump off of a bridge with a bungee cord; as I start to bounce up and down on that cord, this large force is being balanced and even overcompensated by the bungee cord. Similarly if I have a capacitor between the two conductors, then before that capacitor is filled all the way up, there is a little current through the capacitor which is being fought by the force of the capacitor. Or if I connect the two terminals with an inductor, there is a large resistance at first from the inductor because it wants to keep a constant current running through itself and that current is starting at zero.
Maybe the most conspicuous place that this happens is in fluids. It is absolutely routine to go to the doctor and have your blood pressure checked; measuring pressure is very easy. But right after a heart surgery, like putting in a stent, it might be a mistake to say “the heart is at higher pressure than the part of the artery downstream of my stent, I guess blood must be flowing from the heart to that part of the artery.” We are talking about a complex nonlinear system and it is possible that the artery now has blood flowing backwards through it, which could potentially be a fatal medical condition. So then you start to think about “what could I use to actually measure fluid flow through this thin channel?” and some of the options are to, say, inject saline into the artery and detect it downstream in a “time of flight” measurement or else to try to fire ultrasound or light both upstream and downstream through the artery in a “doppler shift” measurement. But just wrapping something around the vessel and measuring pressure cannot give you the information you want except under assumptions that the system is very simple—we might say, “linear.”
A: It's true that

the voltage can be increased and the current decreased [while] keeping the power constant,

but as Alfred Centauri said, this is not particularly natural to do in a single wire, because in order to keep the power constant you need to simultaneously adjust the resistance to remain proportional to the square of the voltage.
Let's say you do that - you double the voltage and quadruple the resistance. Then intuitively, the doubled voltage means that you're pushing electrons through the wire twice as hard. (More precisely, you're subjecting them to twice as strong of an electric field, since the voltage is proportional to the average electric field along the wire.) But since you've quadrupled the resistance, the wire is four times as "viscous" as before, so it's harder to push electrons through it. Since you're pushing twice as hard through a medium that's four times as resistive, the net result is that only half as many electrons pass through (per second) as before. So the current goes down.
Since you have to work twice as hard in the second situation to push each individual electron through the wire, each electron carries twice as much energy as before. But since you're only pushing half as many electrons through as in the first situation, the total work you have to expend is that same in both situations. In the second situation, you're distributing the same total amount of energy across fewer electrons, so that each individual electron carries more energy.
A: Power is the rate of doing work, Joules per second 
Voltage is the work done per unit charge to move the charge between two points. Joules per coulomb.
Current is the rate of movement of charge between the two point. Coulombs per second.
Voltage times current is therefore the work done per second to move charge between two points.  Joules per second (power)
If I do more electrical work per unit charge to move the charge between two points  (increase the voltage) I can proportionately move less charge between the two points per second (reduce the current) and produce the same power.
Consider a mechanical analogue.
$$P=F\frac{d}{s}=Fv$$
Consider voltage being analogous to force and velocity analogous to current. The greater the force (voltage) the less the velocity (current) you need to produce the same mechanical power. Of course force and velocity are not the same as voltage and current, but perhaps the analogy will give you a more intuitive feeling for electrical power.
Hope this helps 
A: 
Considering that voltage is a measure of tendency of a terminal to
  send or receive electrons, how can we have a high tendency to do so
  (high voltage), but low rate of passage of electrons (low current)?

But voltage isn't a measure of tendency of a terminal to send or receive electrons. If you think of voltage in this way, it will be very difficult to develop an intuition for electric circuits.
It's important to realize that, for an ideal voltage source, the current through is entirely dependent on the attached circuit. Put another way, an ideal voltage source can source (or sink) as much current as required by the attached circuit independent of the voltage of the source.
Consider a simple circuit with a voltage source producing a voltage $V_S$ connected to a resistor with resistance $R$.
The current $I_S$ produced by the voltage source is
$$I_S = \frac{V_S}{R}$$
and the power $P_S$ delivered by the voltage source is
$$P_S = V_S\cdot I_S = \frac{V^2_S}{R}$$
and so, if the resistance changes, the current and power changes accordingly while the voltage across is fixed.

Now, if we imagine that the voltage source is adjustable and that it is controlled in such a way that the power delivered is the constant $P_0$, then it follows that the adjustable voltage is given by
$$V_S(R) = \sqrt{P_S\cdot R}$$
That is, for the voltage source to produce constant power, the source voltage must be proportional to the square root of the connected resistance. Note that this isn't a typical application of a voltage source.
A: The question needs to be clarified: The important idea of “high voltage low current” almost certainly will be the situation for mains electricity supply to our houses etc. (it hardly makes any sense in other situations) . The point being that our house needs to use a certain fixed power $P=VI$. Now how best to get that power transferred to us from the generating station 100 km away where we have cables which need to be not too heavy and not lose a lot of power heating the cables instead of heating our house. For high voltage low current we don’t need to worry so much about losing energy heating the cables.
An alternative say lower voltage higher current will inevitably result in more joule heating losses in supply cables  unless you start making cables thicker to reduce their resistance but that is not wanted as they would cost more and be heavier and more tricky to deploy hanging from pylons etc. Wikipedia explains more I am sure.
