List indexing starts at zero, so for example a fourth element in a list has index 3. You can define a helper function to get items/columns by their "actual position".

def column(matrix, i):
    return [row[i-1] for row in matrix]
column(table,2)

Out[15]:
['B', 'F', 'J', 'N', 'R', 'V', 'Z', '4', '8']

def getitem(matrix,row,column):
    return matrix[row-1][column-1]
getitem(table,2,3)

Out[16]:
'G'

As for your edit, table[1][2] should print G, not O and table[4][3] rightly returns T.

Try this:

matrix = "\n".join([" ".join(["0" for x in range(4)]) for x in range(4)])
print matrix

The thing is that you're trying to print a list object instead of a string. This way you join all of the numbers into 1 large string separated with spaces for columns and \n (line endings) for rows.

Keep in mind this method will not work for 2 digit numbers unless you pad the 1 digit with a leading zero or space. In order for it to work in all cases, I suggest using the Tabulate module.

You could try:

for i in range(a[0]):
  for j in range(a[1]): 
    if (j == a[1]-1):
      print(x[i][j])
    else:
      print(x[i][j], end = " ")

You could also simplify your code a little bit to:

a=[int(n) for n in input().split()] # e.g. a = [2,4]

matrix = [[i*a[1]+j+1 for j in range(a[1])] for i in range(a[0])]
for m in matrix:
    print(*m)