One-time pad using RSA and Diffie-Hellman functions Suppose that Alice and Bob use an encryption algorithm based on a one-way function $F$ (Eve knows it), and $r$, a secret key shared between Alice and Bob (Eve doesn't know it).
Key is computed as: $K = F(r)||F(2r)||F(3r)||...$, and the ciphertext is computed as $C = M \oplus K$, and all the blocks of $F(r)$ are $1024$-bit long.

If $F(x) = x^e \mod n$, and $(n, e)$ pair is public, will the first $1024$ bits of both plaintext and ciphertext be enough for Eve to read the entire message?

Same for $F(x) = g^x \mod p$, and public $(g, p)$ pair.

Also, are the cases described above good examples of public-key cryptography? But Eve can't encrypt only having public keys, as she doesn't have $r$. So interception of $1024$-bit long plain/cipher pair is the only way for her to obtain some information from the system.
This feels like homework, and so I won't give you the answer; I'll give you hints:


  If F(x) = x^e mod n, and (n, e) pair is public, will the first 1024 bits of both plaintext and ciphertext be enough for Eve to read the entire message?


If Eve knows the first plaintext $M_0$, and the ciphertext $M_0 \oplus (r^e \bmod n)$, how can Eve recover $r^e \bmod n = F(r)$?  Given that, how can she reconstruct $F(kr) = (rk)^e \bmod n = r^e \cdot k^e \bmod n$?


  Same for F(x) = g^x mod p, and public (g, p) pair.


Again, given $F(r) = g^r \bmod p$, how can Eve recover $F(kr) = g^{kr} \bmod p = (g^r)^k \bmod p$?


  Are the cases described above good examples of public-key cryptography?


Nope; they're not examples of public-key cryptography as all.  We know this based on your description:


  Suppose that Alice and Bob use an encryption algorithm based on ... r, a secret key shared between Alice and Bob


Public key cryptography doesn't use a secret key shared between the two parties.

Also, please don't describe these algorithms as 'one-time-pad'.  That has a very specific meaning (where the pad is generated in a truly random fashion, and not algorithmicly); this is not that.