Known Plaintext attacks using standard greetings etc In all modern ciphers, even if one bit of input changes, then half of the output will change because of diffusion. Considering this, how exactly will knowing standard salutations, etc. (&quot;Hello Alice&quot; or other known parts of the message) be used to launch an attack. You don't know the whole PlainText but only part of the PlainText. So how exactly will it help?
Known Plaintext Attack (KPA) is the security level that is below of Chosen Plaintext Attack (CPA) that we want at least CPA from all ciphers. Actually, we want more in the modern cryptography; Ind-CPA.
The classical cryptography era contains lots of examples that can be easily broken with KPA attack; shift, permutation, Vegenere, and Hill ciphers are some examples.
In the modern era, the KPA can be used for key searching as below.

Brute-force attack; search for all keys

  # c = E(k,m) for some key k we are looking for
  for i in keys:
    if c == E(i,m):
       return i

An interesting example in the history was the RSA's DES challenges, which contains given partial information The unknown message is: on the full plaintext and the contestants are required to find the key given the full ciphertext. They used brute-force attacks similar to the above. The attack is successful since the key size of DES 56-bit.
This above can be still considered a classical way to find the key and the technological advances help to find the key. A modern novel attack is coming from Matsui and Yamagish in 1992 on their novel paper

A new method for known plaintext attack of FEAL cipher

This is now called the Lineart Attack. On attacking the DES, it requires $2^{43}$ known plaintexts by Matsui in 1994.

Linear Cryptanalysis Method for DES Cipher

If you want to learn about the Linear attack the tutorial by Heys is a good starting point.
Now back your specific question;

(&quot;Hello Alice&quot; or other known parts of the message) be used to launch an attack. You don't know the whole PlainText but only part of the PlainText. So how exactly will it help?

First of all, with the Kerckhoff's principle, we assume that except the key we assume all known about the target system, like how the messages turned into bytes for encryption.

If the position is known as in the RSA DES challenges, then this knowledge can be easily turned into a known-plaintext if fits into one block. One needs at least 16 characters for AES and 8 for DES.
The Hello Alice is forming more than one block for DES Hello Al and ice?????. If you brute-force the DES then the first block can be used to find the key candidates, yes there be more than one candidate. Then the decryption of the next blocks can be used to pinpoint the messages.
# c = E(k,m) for some key k 
# we are looking for the possibele key for full known message block m
for i in keys:
   if c == E(i,m):
      m' = D(i,c')             # c' is the other message blocks  
      if m' is a valid string  # check the validity of m'
        return i 


A special case if we don't have a full block, then execute the brute force by decryption, and for each possible candidate, which contains the partial information, look for the meaningful strings by decrypting the rest ciphertexts. This can be automized with tools like Linux's string command and possibly check the PKCS#5 padding at the end will eliminate most of the candidates.


If the position is not known
This is a little problem to be solved compared to brute force and really depends on the message size. If the message size is $t$ and the known message part is $l$ then one needs to look for $t-l$ positions. Therefore the cost of brute force was multiplied with $t-l$. For short messages, this is not a real problem, however when the message size is more than $2^{32}$ then the brute-force cost become $2^{80}$ for DES and $2^{160}$ for AES.