ATM90E26 Calibration method / circuit
But to get a power factor equal to 0.5 I'm no really sure how.

Start by examining the power triangle: -

Then define how much current you want to flow. As an example, I've chosen 0.5 amps. That means that the impedance required (the hypotenuse on the power triangle) is 220 volts divided by 0.5 amps = 440 Ω.
That impedance is made up from a resistor in series with an inductor and together they form the right angle in the power triangle above.
Given that \$\Phi\$ is 60° for a power factor of 0.5, reactance (\$X\$) can be found by re-arranging this: -
$$ \sin(60°) = \dfrac{X}{Z}\text{ or } X = Z\cdot\sin(30°) = 381.05\text{ Ω}$$
And, using the cosine relationship for sides and angles, R = 220 Ω.
So, if your mains AC frequency is 50 Hz, the inductance required to be in series with the 381.05 Ω resistor is: -
$$L = \dfrac{381.05}{2\cdot\pi\cdot 50} = 1.2129 \text{ henries}$$

So here comes the question, how can I make a test / calibration
circuit that generates a power factor equal to 0.5? I mean with what
type of every day use device can I achieve this? will a common
fan(220V AC) do the trick?

Well, to most EEs making an inductor can be done from commonplace stuff but if you mean making it from stuff that anyone has common access to my advice is forget it. Do the job properly if you want an accurate calibration.