Regarding miniature solar panels
I dont understand in which irradiance does the panel give those values. This is my first question.


Not too hard to calculate from the data provided, given a few assumptions. The "peak voltage" they give is less than the open circuit voltage, so ostensibly they mean to specify the peak voltage and current at the maximum power point. The "mp" after "Imp" and "Vmp" would also conventionally indicate "maximum power". So they are saying this panel can produce (under optimal conditions):

$$ 3 \mathrm V \cdot 0.2 \mathrm A = 0.6 \mathrm W $$

The area of the panel is:

$$ 92 \mathrm{mm} \cdot 61 \mathrm{mm} = 0.005612 \mathrm m^2 $$

They claim an efficiency of 15%, so the irradiance \$I\$ must have been:

$$ 0.15 \cdot I = \frac{0.6 \mathrm W}{0.005612 \mathrm m^2} \\
I = 713 \mathrm W / \mathrm m^2 $$

Irradiance on a very clear day at Earth's surface with the sun directly overhead is on the order of 1kW per square meter, so this sounds reasonable enough, considering that the efficiency number may not account for the fact that this device has some margins around the panels.


  My second question is why there is no datasheet? I guess that depending on the type of panel (for example polycrystaline) the I-V curves are same for every panel. Am I right?


Some products just suck. That said, silicon junction photovoltaics pretty much have a common shape to their I-V curves, which is more or less a top-for-bottom flip of the I-V curve of a silicon diode. A more expensive product would probably have more detailed specifications; for something cheap like this you will just have to infer or measure for yourself.