Explanation of Lipo Cut off Circuit
"1025" in your formula above is an error due to the shocking diagram quality (drawn by me :-( ).

It should be 1.25 as the TLV431 regulator output = 1.25V. 

So to achieve 6.2V cutoff you want
R2 = R3 x (6.2/1.25 -1 )
= R3 x (4.96-1)
=  3.96 x R3 ~+ 4 x R3.   

If R3 = 10k then R2 = 39K is OK.   

IC2 is any opamp that will work on the available power supply voltage and which has enough output swing to drive the MOSFET off and on.  

The 2nd circuit using a transistor instead of an opamp will work as well and saves you needing to use an opamp. A resistor from MOSFET gate to ground would be  auseful precaution to ensure it is off when not driven on - anything in the 10k-100k range would work OK.
The transistor can be almost any small PNP. I'd use a BC327 if through hole or a BC807 if SMD BUT almost any PNP transistor will work.
Using R1 = 39k and R2 = 10k should work OK.

Note that this circuit draws a small current via R1+R2 when the battery is cut off.
Iquiescent ~+ Vbat/(R1+R2) = 6.2 / (49k) =~ 125 uA.
 This will often not be a problem.