Several questions about Paillier cryptosystem
Let us briefly recall the Paillier encryption.
Let $k_{pub} = (N = PQ, g)$ be a public key, where $N$ is the RSA modulus.
The secret key is $\lambda = \mathrm{lcm}(P-1,Q-1)$ (or $P,Q$).
The encryption of $p \in \mathbb{Z}_N$ with randomness $r \in \mathbb{Z}_N^*$ is $C = g^p r^N \bmod{N^2}$.

You can verify $\mathbb{Z}_{N^2}^* \simeq \mathbb{Z}_N \times \mathbb{Z}_N^*$.
As Paillier wrote, there is a subgroup $G = \{z \mid z = y^N \bmod{N^2}\}$ of order $\phi = (P-1)(Q-1)$.
There is a bijective mapping $\tau$ from $\mathbb{Z}_N^*$ to $G$ which maps $y$ to $z = y^N \bmod{N^2}$. The inversion $\tau^{-1}(z)$ is computed by $y = z^{N^{-1} \bmod{\lambda}} \bmod{N}$. Roughly speaking, this is the RSA decryption with $e = N$.

Answers:


Yes.
You can find such $r$. Since you know the plaintext $p$, you can compute $d = g^{-p}C = r^N \bmod{N^2}$. By applying $\tau^{-1}$, you can find $r$. (See Section 5.)
No. There is only one $r \in \mathbb{Z}_N^*$. (See Lemma 3.)\
Makes sense since there may be even a huge number of ciphertexts representing the same plaintext but they all have to decrypt to just one plaintext. Hence r is unique as long as en- and decryption are executed using the same pair of keys.
No.