For question (1):
This page gives some hints on IVs and CBC: https://defuse.ca/cbcmodeiv.htm
I copy-paste the part about IVs "predictability"
Randomness is not enough, though. IVs have to be unpredictable,
Suppose there is a CBC-mode encryption system that selects a random
IV, publishes it, asks the user for a one-block plaintext to encrypt,
encrypts it with that IV, then gives the ciphertext to the user.
Suppose Alice uses the system to encrypt two distinct messages A, and
B, to get ciphertexts C and D. Alice gives Mallory the plaintexts and
the ciphertexts and offers Mallory \$1000 if he can tell her which of
the two ciphertexts is the encryption of plaintext A. If he can't, he
has to give Alice \$1000.
If Mallory made a random guess, he would be right with 50%
probability, because either C corresponds to A, or D corresponds to A.
If the system is secure, Mallory shouldn't be able to do any better
Mallory doesn't have to guess, though, because he can use a
chosen-plaintext attack on the CBC-mode encryption system to figure
out if C corresponds to A, or D corresponds to A. Mallory knows that
the IV Alice used to encrypt A was IVA, and he knows that the input to
the block cipher was A XOR IVA. Mallory just needs to know whether the
block cipher encryption of A XOR IVA is C or D. Mallory asks the
encryption system for the next IV, IVN, and sends it the plaintext A
XOR IVA XOR IVN to encrypt. The system follows CBC mode, XORing
Mallory's plaintext with IVN and passing the result to the block
cipher. IVN XOR IVN is 0, so the system passes A XOR IVA to the block
cipher, and gives Mallory the ciphertext. What Mallory gets back is
either C or D, whichever one corresponds to plaintext A. In this case,
Mallory gets back C, tells Alice that C corresponds to A, and wins
$1000 with 100% probability.
Mallory would not have been able to do this if he could not predict
the IV, since the plaintext he sends to the system depends on the next
Although the IV must be unpredictable (for each key), it doesn't need to be kept safe once the ciphertext was generated. Generally it is send (prefixed) to the ciphertext. It's certainly not part of the key.
If there are separate encryptions, then see 1. If there is just one long stream then you only need one IV.