a homomorphic operation on Paillier encryption Assume  $N=pq$, where $p$ and $q$ are two strong prime numbers. 

Also, assume we have  finite field $\mathbb{F}_u$ where $u$ is a $112$ bit prime number. Let $r_i$ be a uniformly random element of $\mathbb{F}_u$.

Let $(r_i)^{-1}$ be multiplicative inverse of $r_i$ in ring $Z_N$.

We encrypt a message as $c=E(r\cdot m)$ using Paillier encryption. 



Question 1: Given $c$ and $(r_i)^{-1}$, can we "always" perform $c^{(r_i)^{-1}}=E((r_i)^{-1}\cdot (r_i)\cdot m)$ such that its  decryption value only contain $m$?



Question 2: given $E(m)$,  how can we chose its additive inverse $m'$ such that $E(m).E(m')=E(m+m')$ so its decryption value would be 0? should $m'\in \mathtt{Z}_N$ or $m'\in \mathtt{Z}_N{^2}$
Question 1: Given $c$ and $(r_i)^{-1}$, can we "always" perform $c^{(r_i)^{-1}}=E((r_i)^{-1}\cdot (r_i)\cdot m)$ such that its 
  decryption value only contain $m$?


Yes. You can multiply arbitrary values $x$ into a message using paillier encryption and it will result in multiplication $\bmod n$ and thus you'd get $m$ back.

This would happen by  exponentiating the cipher text with the $x$, as $c^x\bmod {n^2}$ as you described.


  Question 2: given $E(m)$,  how can we chose its additive inverse $m'$ such that $E(m).E(m')=E(m+m')$ so its decryption value would be
  0? should $m'\in \mathtt{Z}_N$ or $m'\in \mathtt{Z}_N{^2}$


I don't think there's any way given a ciphertext of $m$ to construct a plaintext $m'$ that would yield $m+m'=0$ as this would trivially equal a plaintext recovery and thus a (nearly) total break of the encryption.

However, you can use the answer from question one to calculate the additive inverse ciphertext, which would be $c^{n-1}\bmod{n^2}$. If you multiply this with the original cipher text, you will get an encryption of $0$, to which you can add whatever you want (e.g. an encryption of 1?).

Alternatively you might as well just exponentiate the ciphertext with $0$ to get the same result...