21v solar panel safety issues, and how to get a consistent 12v from it
The panel is "safe enough" when used with normal precautions you should use with any battery powered experiments or equipment.


The 5V USB outputs will allow charging and operation of 5V equipment. 
The "12V" (18V / 21V see below) output will allow charging of 12V lead acid batteries (car / motorcycle / alarm) and operation of 12V equipment that is tolerant of up to about 18V when loaded and somewhat more under light loads.   
Use of a 12V battery (such as an old car battery) will allow the 12V output to be at about 12V when experimenting.
The 12V output should provide ABOUT 2.75A at 18V in full sun. It will provide maybe 3A when loaded to 12V or below in full sun.
ANY energy supply system has some risks. s long as these are understood and allowed for they become part of learning. At 50 W and 18V (see below) the output COULD cause a fire (it may take some effort). 
Shock is unlikely
Do not put wires in mouth or stand in metal bucket of salt water with one lead attached to bucket and holding other lead !!! :-) . 


I say that to make the point that it is very safe usually BUT that if you tried the bucket experiment you'd strongly notice the result.
You'd regret putting both wires in your mouth! - but you'd about equally regret it if a car battery was the energy source. ]. 





Your PV panel consists of 35 PV cells in series (5 columns of 7 cells) - see image below.
These will produce ABOUT 0.5V/cell in full sunlight when loaded or 35 x 0.5 = 17.5V - say around 18V when fully loaded.   

Optimally loaded for maximum power values in full sunlight under optimum loading are said to be at the "maximum power" point and suffixed mp.

So here Vmp = 18V. 

You are seeing 21V O/C (open circuit) so Vmp/Voc = 18/21 ~= 85%.
This is consistent with what would be expected.  

A "12V"panel is typically about 18Vmp loaded so that it can fully charge a 12V lead acid battery (about 14,4 V) plus a diode drop (about 1V at full current) + wiring and connection losses.

At 50 Watt (Wmp) the current (Imp) will be I = VW/V  Wmp/Vmp = 50/18 = 2.78 A.
This is under full sun conditions at noon with the panel optimally aligned to the sun (point panel at sun and angle panel to maximise shadow length)

The 5V outputs are very hopefully provided via a DC-DC converter so should be capable of currents up to 10A total. However, they may have cut corners and provided a less capable converter. Experimenting will tell. This SHOULD charge cellphones and similar with ease BUT I'd be wary of committing an expensive device to the mercy of a panel of unknown provenance and quality - if the 5V output voltage rose to 12V, or 18V, or 21V a connected cellphone or other device would quite probably be destroyed. If using for 5V charging I'd consider the use of an external  charger that I was confident was trustable.