First thing is to remember that python uses zero indexing.

You can iterate throught the list except using the range function to get the indexes of the items you want or slices to get the elements.

What I think is becoming confusing here is that in your example, the values and the indexes are the same so to clarify I'll use this list as example:

I = ['a', 'b', 'c', 'd', 'e']
nI = len(I) # 5

The range function will allow you to iterate through the indexes:

for i in range(1, nI):
    print(i)
# Prints:
# 1
# 2
# 3
# 4

If you want to access the values using the range function you should do it like this:

for index in range(1, nI):
    i = I[index]
    print(i)
# Prints:
# b
# c
# d
# e

You can also use array slicing to do that and you don't even need nI. Array slicing returns a new array with your slice.
The slice is done with the_list_reference[start:end:steps] where all three parameters are optional and:
start is the index of the first to be included in the slice
end is the index of the first element to be excluded from the slice
steps is how many steps for each next index starting from (as expected) the start (if steps is 2 and start with 1 it gets every odd index).
Example:

for i in I[1:]:
    print(i)
# Prints:
# b
# c
# d
# e

My 5 cent:

start_from = 'b'

for val in l[l.index(start_from ) if start_from  in l else 0:]:
   print val

Just slice it.

for script, location in self.device.scripts[1:]:
    pass

For your second question, you don't need to worry about any IndexError since slicing returns an empty list when it's out of range.

You can explicitly create an iterable with the iter() builtin, then call `next(iterable) to consume one item. Final result is something like this:

line_iter = iter(list_of_lines[:])
# consume first item from iterable
next(line_iter)
for index, item in enumerate(line_iter, start=1):
    list_of_lines[index][1:3] = [''.join(item[1:3])]

Note the slice on the first line, in general it's a bad idea to mutate the thing you're iterating over, so the slice just clones the list before constructing the iterator, so the original list_of_lines can be safely mutated.

islice has the advantage that it doesn't need to copy part of the list

from itertools import islice
for day in islice(days, 1, None):
    ...

If you want to get the second column of an array, you could use a list comprehension, like so:

a = [ [ 1, 2, 3, 4 ],
      [ 5, 6, 7, 8 ],
      [ 9,10,11,12 ],
      [13,14,15,16 ] ]


second_column = [ row[1] for row in a ]
# you get [2, 6, 10, 14]

You can wrap this up in a function:

def get_column ( array, column_number ):
    try:
        return [row[column_number] for row in array]
    except IndexError:
        print ("Not enough columns!")
        raise # Raise the exception again as we haven't dealt with the issue.

fourth_column = get_column(a,3)
# you get [4, 8, 12, 16]

tenth_column = get_column(a,9)
# You requested the tenth column of a 4-column array, so you get the "not enough columns!" message.

Though really, if you're working with rectangular arrays of numbers, you want to be using numpy arrays, not lists of lists of numbers.

Or, by Lattyware's implied request, a generator version:

def column_iterator ( array, column_number ):
    try:
        for row in array:
            yield row[column_number]
    except IndexError:
        print ("Not enough columns!")
        raise

Usage is just like a normal list:

>>> for item in column_iterator(a,1):
...    print(item)
... 
2
6
10
14
>>> 

The generator-nature is evident by:

>>> b = column_iterator(a,1)
>>> b.next()
2
>>> b.next()
6
>>> b.next()
10
>>> b.next()
14