If you print the value of words[i] after your attempted character replacements you will see that it is set to a tuple, e.g.

('word', (',', ''), ('/', ''), ('?', ''), ('!', ''))

So the line that tries to remove unwanted punctuation actually creates a tuple because that's what the comma separated items are, i.e.

words[i].replace(".", ""), (",", ""), ("/", ""), ("?", ""), ("!", "")

is actually a tuple consisting of words[i].replace(".", "") followed by (",", ""), etc.

You might have meant to chain a whole lot of replace operations together, but that would need to look like this:

words[i].replace(".", "").replace(",", "").replace("/", "").replace("?", "").replace("!", "")

But that is pretty ugly, and it's restricted to just a few punctuation symbols. str.translate() is better:

words[i] = words[i].translate(None, '.,/?!')

or, if you want to get rid of all punctuation you can use string.punctuation:

import string
words[i] = words[i].translate(None, string.punctuation)

Or, if you are using Python 3:

import string
words[i] = words[i].({ord(c):None for c in string.punctuation})

There are other problems in your code, but see if you can correct this first issue first.

Cell coordinates should be provided as a string:

ws['A1'] = 'testing 1-2-3'

Or, if you want to use row and column indexes, use ws.cell().value:

ws.cell(row=1, column=1).value = 'testing 1-2-3'

You return four variables s1,s2,s3,s4 and receive them using a single variable obj. This is what is called a tuple, obj is associated with 4 values, the values of s1,s2,s3,s4. So, use index as you use in a list to get the value you want, in order.

Copyobj=list_benefits()
print obj[0] + " is a benefit of functions!"
print obj[1] + " is a benefit of functions!"
print obj[2] + " is a benefit of functions!"
print obj[3] + " is a benefit of functions!"

tagged begins as a list of pairs of tuples. So, word and tag are both tuples. e.g. the first values for word and tag will be

word : ('no', 'DT') 
tag : ('coment', 'NN')

Perhaps, try:

temp = []
    for x in tagged: 
        for y in x: 
            temp.append(y) 
tagged = temp

before running the loop - assuming you want to iterate over the tuples in your main loop.

This error indicates that value is a tuple, and not a string as you might expect. This indicates a problem with your application.

Here the problem is that fetchone() returns a one-tuple. You should change from this line:

value = results.fetchone()

to this (notice the comma after value):

value, = results.fetchone()

or this (not recommended):

value = results.fetchone()[0]

But why is fetchone() returning a tuple instead of a string? Because you can SELECT multiple columns. Consider for example the following SQL statement:

SELECT a, b, c FROM my_table;

In this case, fetchone() will return a three-tuple.

Writing value, = fetchone() you are telling Python that you are expecting a one-tuple and you want that single item placed into value. If you were expecting the three-tuple, you'd have used column_a, column_b, column_c = resulsts.fetchone().

This is the reason why you should prefer value, over fetchone()[0].

Bonus tip: I noticed you are using Python 3. In this case, you can write:

translation_table = dict.fromkeys(b'+-(),', None)

Speeding up your code and making it more clean.