cant understand the code.

Basically, if you number the 8 bits in a byte from 0 (least significant) to 7 (most significant), the method tells you whether the bit at position 'pos' is set or not.

(byte)(1 << pos) gives you a byte in which only the bit at position 'pos' is set. If you then '&' this byte with the actual byte, b, you mask off all the other bits i.e. set them to zero. 

If you're left with something which is non-zero, the bit you're interested in must be set (true) but if it's zero then it isn't set (false).

<< this is a left shift operator 
.
The leftshift operator is the equivalent of moving all the bits of a number a specified number of places to the left:

private bool GetBit(byte b, int pos)
{
return ((b & (byte)(1 << pos)) != 0);
}

here you will pass value for b, and pos

you can about this better from this link
http://www.c4learn.com/c-programming/c-bitwise-left-shift-operator/

after checking all condition it will return a value.

Suppose b = 0x05 and pos=2

'1' 00000001
shift '1' 2 positions gives 00000100

shifted '1' 00000100
b           00000101
&           --------
result      00000100

Check if result is zero; if not, return true, else return false