I added an example to reproduce the bug.
From the command line the same code:
Python 2.7.1+ (r271:86832, Apr 11 2011, 18:05:24) [GCC 4.5.2] on linux2

$ python sort_test.py

Everything fine.

Python 3.2 (r32:88445, Mar 25 2011, 19:28:28) [GCC 4.5.2] on linux2

$ python3 sort_test.py 
Traceback (most recent call last):
  File "sort_test.py", line 1821, in <module>
    r = sorted(d.items(), key=operator.itemgetter(1), reverse=1)
TypeError: unorderable types: dict() < dict()

This is expected behaviour:  Python 3 changed the semantics of the comparison operators <, <=, >, >=.  See:

http://docs.python.org/py3k/whatsnew/3.0.html#ordering-comparisons

for more.

I am aware of this change.
In this example, I'm sort by item number 1, which is a list, and its first value is an int.

$ python3 sort_test.py 
<class 'list'>
<class 'list'>
<class 'list'>
<class 'list'>
<class 'list'>
<class 'list'>
...

Dict index is always No. 2. But I do not sort it.
That's why it surprised me this error because nowhere dicts should not be compared.

Sure. I know what's going on.
Sorry for the inconvenience.

> In this example, I'm sort by item number 1, which is a list, and its
> first value is an int.

?  You're sorting by the values of the dict d, and those values have the form [int, int, dict];  so when the two ints match (e.g., in your data, there are two values of the form [64, 124, {...}]) there's a dictionary comparison.

Did you mean to do:

    sorted(d.values(), key=operator.itemgetter(1), reverse=1)

?

I am interested in sorting only by INT0, in this example:
{k: [INT0, INT1, DICT], k: [INT0, INT1, DICT], ...}
not cmp. whole lists.

Unfortunately I can not take advantage of .values() as the keys I need.