Well, you could do this:

>>> if all(k in foo for k in ("foo","bar")):
...     print "They're there!"
...
They're there!

Use the builtin function all()

>>> d = { 'a': 1, 'b': 2, 'c': 3 }
>>> keys = ('a', 'b')
>>> all(elem in d for elem in keys)
True
>>> keys = ('a', 'b', 'd')
>>> all(elem in d for elem in keys)
False

You can use set intersections:

if not d.viewkeys() & {'amount', 'name'}:
    raise ValueError

In Python 3, that'd be:

if not d.keys() & {'amount', 'name'}:
    raise ValueError

because .keys() returns a dict view by default. Dictionary view objects such as returned by .viewkeys() (and .keys() in Python 3) act as sets and intersection testing is very efficient.

Demo in Python 2.7:

>>> d = {
...   'name': 'name',
...   'date': 'date',
...   'amount': 'amount',
... }
>>> not d.viewkeys() & {'amount', 'name'}
False
>>> del d['name']
>>> not d.viewkeys() & {'amount', 'name'}
False
>>> del d['amount']
>>> not d.viewkeys() & {'amount', 'name'}
True

Note that this tests True only if both keys are missing. If you need your test to pass if either is missing, use:

if not d.viewkeys() >= {'amount', 'name'}:
    raise ValueError

which is False only if both keys are present:

>>> d = {
...   'name': 'name',
...   'date': 'date',
...   'amount': 'amount',
... }
>>> not d.viewkeys() >= {'amount', 'name'}
False
>>> del d['amount']
>>> not d.viewkeys() >= {'amount', 'name'})
True

For a strict comparison (allowing only the two keys, no more, no less), in Python 2, compare the dictionary view against a set:

if d.viewkeys() != {'amount', 'name'}:
    raise ValueError

(So in Python 3 that would be if d.keys() != {'amount', 'name'}).

If I understand correctly, the following function should satisfy your specs.

>>> def get_first(dict_, keys,  default):
...     for k in keys:
...         try:
...             return dict_[k]
...         except KeyError:
...             pass
...     return default
... 
>>> get_first(d, [-1, 0, 3, 6], 'default')
4
>>> get_first(d, [-1, 0], 'default')
'default'

For fun, a recursive variant...

>>> def get_first(dict_, keys, default):
...     keys = iter(keys)
...     try:
...         k = next(keys)
...     except StopIteration:
...         return default
...     return dict_.get(k, get_first(dict_, keys, default))

>>> d = {1:2, 3:4, 5:6}
>>> get_first(d, [-1, 0, 3, 6], 'default')
4
>>> get_first(d, [-1, 0], 'default')
'default'

Alternatively, we can do it shorter via calling next on a genexp. Hashes the existing key twice, though.

>>> next((d[k] for k in [-1, 0, 3, 6] if k in d), 'default')
4
>>> next((d[k] for k in [-1, 0] if k in d), 'default')
'default'

Here's one way to do it without having to use .items():

for obj in listofobjs:
    if 'tel' in obj and 'nam' in obj and obj['tel']==tel and obj['nam']==nam:
        ...

Or you could ask for forgiveness provided all dictionary access in the if block are safe:

 for obj in listofobjs:
    try:
        if obj['tel']==tel and obj['nam']==nam:
            ...
    except KeyError:
        pass