Try this:

data['q3'] = data['q3'].str.replace('[', '').replace(']','')

Use str.replace:

df.columns = df.columns.str.replace("[()]", "_", regex=True)

Sample:

df = pd.DataFrame({'(A)':[1,2,3],
                   '(B)':[4,5,6],
                   'C)':[7,8,9]})

print (df)
   (A)  (B)  C)
0    1    4   7
1    2    5   8
2    3    6   9

df.columns = df.columns.str.replace(r"[()]", "_", regex=True)
print (df)
   _A_  _B_  C_
0    1    4   7
1    2    5   8
2    3    6   9

A period '.' is considered a special regex character. To match a character having special meaning in regex, you need to use a escape sequence prefix with a backslash \.

Otherwise, if you use . (dot) standalone and without a backslash, as you have in your example, the period is going to match any character (except for next line) and then replace it with a - (dash)

The only change is adding a backslash before the period:

df.index = df.index.str.replace('\.','-', regex=True)

replace looks for exact matches (by default unless you pass regex=True but you will need to escape the parentheses - see @piRSquared's answer), you want str.replace:

SF['NewPhone'] = SF['Phone'].str.replace("(",'xxx')

which will replace all occurrences of the passed in string with the new string

Example:

In[20]:
df = pd.DataFrame({'phone':['(999)-63266654']})
df

Out[20]: 
            phone
0  (999)-63266654

In[21]:    
df['phone'].str.replace("(",'xxx')

Out[21]: 
0    xxx999)-63266654
Name: phone, dtype: object

If we try replace then no match occurs:

In[22]:
df['phone'].replace("(",'xxx')

Out[22]: 
0    (999)-63266654
Name: phone, dtype: object

See @piRSquared's answer for how to get replace to work as expected (I don't want to cannibalise his answer)