take the input from one python file and use it to another python file

The commenters are right, but it sounds like you want to know anyway. The problem was already highlighted by @martineau. What is happening is the following:

1. test1.py executes in its own process and opens the zip file.
2. test1.py asks the os to start test2.py
3. test2.py executes in its own process, and has no concept of the zip file that exists in the child process of test1.py

Because they do not know about each other, the easiest option you have in the manner you want to do it (i.e., using os.system()), is to pass information from test1.py to test2.py in the form of a string. One way would be the following:

test1.py:

import zipfile
import os

inputzip = input("Enter the Zip File:")
files = zipfile.ZipFile(inputzip, "r")
name_list = (" ".join([i for i in files.namelist()]))
os.system("python3 test2.py {0}".format(name_list))

and

test2.py:

import zipfile
import sys

print ("The list of files in the zip : ")

for name in sys.argv[1:]:
   print (name)

output:

Enter the Zip File:test.zip
The list of files in the zip : 
test
txt1
txt2

In the end this is pretty messy and as already pointed out by @joelgoldstick, you should just import the method from test2.py into test1.py. For example:

test2.py:

def test2_method(zipfile):
  print ("The list of files in the zip : ")

  for name in zipfile.namelist():
     print (name)

test1.py:

import zipfile

from test2 import test2_method

inputzip = input("Enter the Zip File:")
files = zipfile.ZipFile(inputzip, "r")

test2_method(files)

output:

Enter the Zip File:test.zip
The list of files in the zip : 
test
txt1
txt2

I am currently wondering how easy it would be to have both processes share memory and one to be able to pass the memory reference of the opened zip file object to the other. Don't know if this is even possible >_<. Anyone who could enlighten would be cool.