In the line:

Jobs = ()

you create a tuple. A tuple is immutable and has no methods to add, remove or alter elements. You probably wanted to create a list (lists have an .append-method). To create a list use the square brackets instead of round ones:

Jobs = []

or use the list-"constructor":

Jobs = list()

However some suggestions for your code:

opening a file requires that you close it again. Otherwise Python will keep the file handle as long as it is running. To make it easier there is a context manager for this:

with open('Jobs.txt') as openFile:
    x = 1
    while x != 0:
        Stuff = openFile.readline(x)
        if Stuff != '':
            Jobs.append(Stuff)
        else:
            x = 0

As soon as the context manager finishes the file will be closed automatically, even if an exception occurs.

It's used very rarely but iter accepts two arguments. If you give it two arguments, then it will call the first each iteration and stop as soon as the second argument is encountered. That seems like a perfect fit here:

with open('Jobs.txt') as openFile:
    for Stuff in iter(openFile.readline, ''):
        Jobs.append(Stuff)

I'm not sure if that's actually working like expected because openFile.readline keeps trailing newline characters (\n) so if you want to stop at the first empty line you need for Stuff in iter(openFile.readline, '\n'). (Could also be a windows thingy on my computer, ignore this if you don't have problems!)

This can also be done in two lines, without creating the Jobs before you start the loop:

with open('Jobs.txt') as openFile:
    # you could also use "tuple" instead of "list" here.
    Jobs = list(iter(openFile.readline, ''))  

Besides iter with two arguments you could also use itertools.takewhile:

import itertools
with open('Jobs.txt') as openFile:
    Jobs = list(itertools.takewhile(lambda x: x != '', openFile))

The lambda is a bit slow, if you need it faster you could also use ''.__ne__ or bool (the latter one works because an empty string is considered False):

import itertools
with open('Jobs.txt') as openFile:
    Jobs = list(itertools.takewhile(''.__ne__, openFile))

have created a list in my code..

No you haven't you've create a tuple. Use a list instead:

lst = [x,y]

lst.append('red')

The tuple objects cannot append. Instead, convert to a list using list(), and append, and then convert back, as such:

>>> obj1 = (6, 1, 2, 6, 3)
>>> obj2 = list(obj1) #Convert to list
>>> obj2.append(8)
>>> print obj2
[6, 1, 2, 6, 3, 8]
>>> obj1 = tuple(obj2) #Convert back to tuple
>>> print obj1
(6, 1, 2, 6, 3, 8)

Hope this helps!

Tuple is imutable data type means you can not change it. (with some exception) One thing you can do is change contents = () to contents = []

Tuples are immutable meaning they can't be changed unless reassigned. You can however add tuples to a list because lists are mutable, meaning they can be changed. You need to edit the order of your append.

listFollowedBy.append(row)

As of now, your row is a tuple. Tuples don't have an append method, lists do. Why are you appending a list to a tuple instead of a tuple to a list? You've got the method call backwards.

that's because the key creation right term is a tuple

 rests_dict[key] = data_name,data_number

you'd need

if key in rests_dict:
    rests_dict[key].append((data_name,data_number))
else:
    rests_dict[key] = [(data_name,data_number)]

But the best way would be to create

rests_dict = collections.defaultdict(list)  # import collections first

in the first place so no need to test if key exists, just append:

rests_dict[key].append((data_name,data_number))

if key doesn't exist, an empty list is created instead of KeyError