Electrical – SD Card pins dimensions

Your task is certainly not impossible, but I think you would have 90+ hour weeks for those three weeks to get something you might be able to use.

There isn't a single "ZOMG" part of the design; there are a number of smaller blocks that are pretty compartmentalized and straightforward on their own. For example, there's the physical interface: SD cards are essentially (up to) "four-lane" SPI buses, and you can limit yourself to SPI mode to start to make things easier. This isn't terribly complex to do, especially if you have a SPI slave peripheral in hardware that you can use. Supporting enough of the SD protocol to fake out the Windows side probably won't be difficult either, especially if you can limit yourself to emulating an MMC card instead, which is less complex. Emulating a file system is also not terribly difficult. I think where things will get hairy is in the details of interconnecting these blocks and then spending a LOT of time working out the odd little thing that doesn't work.

Like the saying goes... the devil's in the details. From a high level perspective this is doable, although I think your timeframe is unrealistic. Nobody can work 90h weeks and maintain focus and clarity of thought.

And like my favourite saying goes... The difference between theory and reality is that in theory, there is no difference.

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.