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Given just $a$ and $M(a)$, you can construct the message $a\,\|\,b$, where $b = a \oplus M(a)$, which gives the same CBC-MAC output $M(a\,\|\,b) = M(a)$ as $a$ itself. Indeed, we can append the block …

Yes, this is exactly what a message authentication code is for. Its job is to prevent an attacker from tampering with your message, or from forging completely bogus messages. For a secure MAC, it sh …

Note that re-encrypting the output of CBC-MAC with the same key as it was calculated with is equivalent to appending a single all-zero cipher block to the message: $$E_K(\text{CBC-MAC}_K(m)) = E_K(\te …

Yes. Assume that the attacker knows the ciphertext $c = c_1 \mathbin\| c_2$, the initialization vector $v$ and the plaintext $m = m_1 \mathbin\| m_2$. This tells them that $D_k(c_1) = m_1 \oplus v …

IMO, you code looks pretty solid. A few things I might suggest taking a closer look at are: You haven't specified what iteration count you're using for PBKDF2. You should make the iteration count …