Why output is b 2 and not b 1 ?

Question

try:
s = ['a','c']
n = [0,2]
s[1:1] = 'b'
n[1:1] = 1
except:
pass
print(s[1], n[1])

Answer 1

because string is a sequence type, and integer is not. You cannot insert not sequence type in the list (using slicing method).

try:
s = ['a','c']
s is list of two elements 'a', 'c'

n = [0,2]
n is list of two elements 0,2

s[1:1] = 'b'
insert string 'b' to position with index #1
string 'b' is a sequence type, it has 1 element = ['b'],
so it can be inserted to s without any problem
s now contains 3 elements: ['a', 'b', 'c']

n[1:1] = 1
here exception is thrown, as 1 is integer (not a sequence type)
and can't be inserted
list n remains unchanged
it contains [0, 2]

except:
pass
pass do nothing (exception doesn't printed, as try{}except presents )

print(s[1], n[1])
prints second elements of the ['a', 'b', 'c'] and [0, 2]

if you want to get output b 1,
you should replace line
n[1:1] = 1
with line
n[1:1] = [1]

Answer 2

'b', although only one letter, is an iterable. 1, on the other hand, is not.

The slicing method that is used in line 3 and 4 needs an iterable on the right side, so in line you get an error.