From Dive Into Python:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5

Your last error message indicates that you are still mixing lists and arrays. I'll try to recreate the situation:

Make a list of lists. Finding a sublist works just fine:

In [256]: ll=[[1,2,3],[4,5,6],[7,8,9]]
In [257]: ll.index([4,5,6])
Out[257]: 1

Make an array from it - it's 2d.

In [258]: la=np.array(ll)
In [259]: la
Out[259]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

It does not have an index method

In [260]: la.index([4,5,6])
...
AttributeError: 'numpy.ndarray' object has no attribute 'index'

Make it a list - but we get your ValueError:

In [265]: list(la).index([4,5,6])
...
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

That's because list(la) returns a list of arrays, and arrays produce multiple values in == expressions:

In [266]: list(la)
Out[266]: [array([1, 2, 3]), array([4, 5, 6]), array([7, 8, 9])]

The correct way to produce a list from an array is tolist, which returns the original ll list of lists:

In [267]: la.tolist().index([4,5,6])
Out[267]: 1

As for your first question: "if item is in my_list:" is perfectly fine and should work if item equals one of the elements inside my_list. The item must exactly match an item in the list. For instance, "abc" and "ABC" do not match. Floating point values in particular may suffer from inaccuracy. For instance, 1 - 1/3 != 2/3.

As for your second question: There's actually several possible ways if "finding" things in lists.

Checking if something is inside

This is the use case you describe: Checking whether something is inside a list or not. As you know, you can use the in operator for that:

3 in [1, 2, 3] # => True

Filtering a collection

That is, finding all elements in a sequence that meet a certain condition. You can use list comprehension or generator expressions for that:

matches = [x for x in lst if fulfills_some_condition(x)]
matches = (x for x in lst if x > 6)

The latter will return a generator which you can imagine as a sort of lazy list that will only be built as soon as you iterate through it. By the way, the first one is exactly equivalent to

matches = filter(fulfills_some_condition, lst)

in Python 2. Here you can see higher-order functions at work. In Python 3, filter doesn't return a list, but a generator-like object.

Finding the first occurrence

If you only want the first thing that matches a condition (but you don't know what it is yet), it's fine to use a for loop (possibly using the else clause as well, which is not really well-known). You can also use

next(x for x in lst if ...)

which will return the first match or raise a StopIteration if none is found. Alternatively, you can use

next((x for x in lst if ...), [default value])

Finding the location of an item

For lists, there's also the index method that can sometimes be useful if you want to know where a certain element is in the list:

[1,2,3].index(2) # => 1
[1,2,3].index(4) # => ValueError

However, note that if you have duplicates, .index always returns the lowest index:......

[1,2,3,2].index(2) # => 1

If there are duplicates and you want all the indexes then you can use enumerate() instead:

[i for i,x in enumerate([1,2,3,2]) if x==2] # => [1, 3]

Assuming you mean "list" where you say "array", you can do

if item in my_list:
    # whatever

This works for any collection, not just for lists. For dictionaries, it checks whether the given key is present in the dictionary.