Let $(R, \mathfrak m,k)$ be a Noetherian local ring such that the residue field $k$ is infinite. Let $n=\mu(\mathfrak m)$. Then $n=\dim_k(\mathfrak m/\mathfrak m^2)$ . By fixing $x_1,...,x_n \in \mathfrak m$ such that $\bar x_1,...,\bar x_n\in \mathfrak m/\mathfrak m^2$ gives a $k$-vector space basis, we can identify $\mathfrak m/\mathfrak m^2$ with $\mathbb A^n(k)$ by sending $\bar x_i \to e_i$. So we can transfer the classical Zariski topology of $\mathbb A^n(k)$ to $\mathfrak m/\mathfrak m^2$. 

I have two similar kind of questions : 

(1) If $I$ is $\mathfrak m$-primary i.e. $\sqrt I=\mathfrak m$ and the set $\{\bar x\in \mathfrak m/\mathfrak m^2: x\in \mathfrak m \setminus \mathfrak m^2 \space \text{and}  \space (\mathfrak mI:x)=I \}$ is non-empty, then how to show that the set is Zariski Open ?

(2)  If $I=\overline I$ is $\mathfrak m$-primary i.e. $\sqrt I=\mathfrak m$ and the set $\{\bar x\in \mathfrak m/\mathfrak m^2: x\in \mathfrak m \setminus \mathfrak m^2 \space \text{and}  \space (\overline{\mathfrak mI}:x)=I \}$ is non-empty, then how to show that the set is Zariski Open ?


Here for an ideal $I$, by $\overline I$ we denote the integral closure of $I$ https://en.m.wikipedia.org/wiki/Integral_closure_of_an_ideal