What is wrong with the comparison a,b == 1,2?

Essentially this apparently weird behavior come from the fact that the right side of your expression is a tuple, the left side is not.

The expected result is achieved with this line, which compares a tuple with a tuple:

(a, b) == (1, 2)

Your expression instead is equivalent to:

(a, b == 1, 2)

Which is a tuple containing a, the comparison between b and 1, and 2.

You can see the different behavior using the dis module to check what python is doing:

import dis

dis.dis("a,b == 1,2")
  1           0 LOAD_NAME                0 (a)
              2 LOAD_NAME                1 (b)
              4 LOAD_CONST               0 (1)
              6 COMPARE_OP               2 (==)
              8 LOAD_CONST               1 (2)
             10 BUILD_TUPLE              3
             12 RETURN_VALUE

 dis.dis("(a,b) == (1,2)")
  1           0 LOAD_NAME                0 (a)
              2 LOAD_NAME                1 (b)
              4 BUILD_TUPLE              2
              6 LOAD_CONST               0 ((1, 2))
              8 COMPARE_OP               2 (==)
             10 RETURN_VALUE

You can see that in the first evaluation python is loading a, then is loading and b then loading the right side element of the comparison (1) and compare the last two loaded elements, then load the second right-element then build a tuple with the results of those operations and returns it.

In the second code instead python loads the left side (operations 0, 2 and 4) , loads the right side (operation 6), compare them and return the value.

You need to explicitly compare the two tuples using parantheses:

a = 1
b = 2
print((a,b) == (1,2))  # True

Right now, you're creating the tuple (a, b == 1, b). That evaluates to (1, False, 2).