I'd say it's a bug of python3.

def u():
    exec("a=2")
    print(locals()['a'])
u()

prints "2".

def u():
    exec("a=2")
    a=2
    print(a)
u()

prints "2".

But

def u():
    exec("a=2")
    print(locals()['a'])
    a=2
u()

fails with

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in u
KeyError: 'a'

--- EDIT --- Another interesting behaviour:

def u():
    a=1
    l=locals()
    exec("a=2")
    print(l)
u()
def u():
    a=1
    l=locals()
    exec("a=2")
    locals()
    print(l)
u()

outputs

{'l': {...}, 'a': 2}
{'l': {...}, 'a': 1}

And also

def u():
    l=locals()
    exec("a=2")
    print(l)
    print(locals())
u()
def u():
    l=locals()
    exec("a=2")
    print(l)
    print(locals())
    a=1
u()

outputs

{'l': {...}, 'a': 2}
{'l': {...}, 'a': 2}
{'l': {...}, 'a': 2}
{'l': {...}}

Apparently, the action of exec on locals is the following:

• If a variable is set within exec and this variable was a local variable, then exec modifies the internal dictionary (the one returned by locals()) and does not return it to its original state. A call to locals() updates the dictionary (as documented in section 2 of python documentation), and the value set within exec is forgotten. The need of calling locals() to update the dictionary is not a bug of python3, because it is documented, but it is not intuitive. Moreover, the fact that modifications of locals within exec don't change the locals of the function is a documented difference with python2 (the documentation says "Pass an explicit locals dictionary if you need to see effects of the code on locals after function exec() returns"), and I prefer the behaviour of python2.
• If a variable is set within exec and this variable did not exist before, then exec modifies the internal dictionary unless the variable is set afterwards. It seems that there is a bug in the way locals() updates the dictionary ; this bug gives access to the value set within exec by calling locals() after exec.