Denote $\mathbb{N}^*=\{1,2,3,...\}, \dot k = \{k,2k,3k,...\}, \mathbb{P}=\{2,3,5,7,11,..\}$ and write $M\leq \mathbb{N}^*$ to denote that $M$ is lcm-closed, i.e. $a,b\in M\Rightarrow \text{lcm}(a,b)\in M$. Then, an equivalent number-theoretic phrasing of Frankl's conjecture is the following $$\bbox[#FFB797,2pt,border: 2px solid red]{\array{\text{If $\{1\}\neq M\leq\mathbb{N}^*$ is finite and contains only square-free numbers, then}\\ \exists p\in \mathbb{P}:|M\cap \dot p|\geq \frac{1}{2}|M|}}$$ The equivalence can be proven constructing from $M$ the union-closed family $\mathcal{F}_M:=\{\{p\in\mathbb{P}:p\ \vert\ m\}:m\in M\}$ of the universe set $\mathbb{P}$ and constructing from a finite union-closed set $\mathcal{F}$ of the universe set $\mathbb{P}$ the set of square-free numbers $M_\mathcal{F}:=\{\prod_{p\in X}p:X\in \mathcal{F}\}$.
With this new formulation, I thought of generalizing it so that we no longer need the square-free condition. These are the two possible generalizations that I've come up with. If $\emptyset\neq M\leq\mathbb{N}^*$ is finite , then
$$\array{\textbf{Conjecture 1. }\ \exists p\in \mathbb{P}:|M\cap \dot p|\geq \cfrac{\nu_p(M)}{1+\nu_p(M)}|M|}\\ \textbf{Conjecture 2. }\ \exists p\in \mathbb{P}:\sum_{m\in M}\nu_p(m)\geq \frac{1}{2}\nu_p(M)|M|$$ where $\nu_p(m)$ is the $p$-adic valuation of $m$ and I denote $\nu_p(M)=\nu_p(\max M)=\max\limits_{m\in M}\nu_p(m)$. Note as well that, now, we don't need to exclude the case $M=\{1\}$ anymore as $\nu_p(\{1\})=0$. I expect that at least one of them has a counterexample. However, I am quite sloppy at finding them. So then my quesion is
Q: Do the previous generalizations have clear counterexamples?
I will now explain the motivation to why I chose those precise coefficients on the right hand side of both conjectural inequalities to show that they aren't arbitrary. Note that on the original Frankl conjecture, we find that for the finite power set $2^U$ the inequality $$|\{X\in 2^U:\alpha\in X\}|\geq 2^{|U|-1}=\frac{1}{2}|2^U|$$ is, in fact, an equality for every $\alpha\in U$. So, in a sense, one expects that the maximal case is as bad as it gets. Now, it isn't hard to see that the equivalent lcm-closed set to the union-closed set $2^{\{p_1,...,p_k\}}$ where $\{p_1,...,p_k\}$ are distinct primes is the set $\text{Div}(P)$ of divisors of $P=\prod_{j=1}^k p_j$ where $P$ is a square-free number. So, if we consider any $M=\text{Div}(m_0)$ for some $m_0\in\mathbb{N}^*$ (not necessarily square-free), one easily sees that $$\Rightarrow |M\cap\dot p|=d(m_0)- d(m_0p^{-\nu_p(m_0)})=(1+\nu_p(m_0))d(m_0p^{-\nu_p(m_0)})-d(m_0p^{-\nu_p(m_0)})$$ $$=\nu_p(m_0) d(m_0p^{-\nu_p(m_0)})=\frac{\nu_p(m_0)}{1+\nu_p(m_0)}d(m_0)= \cfrac{\nu_p(M)}{1+\nu_p(M)}|M|$$ $$\Rightarrow \sum_{m\in M}\nu_p(m)=\sum_{d\vert m_0}\nu_p(d)=\sum_{j=0}^{\nu_p(m_0)}\sum_{d\vert m_0\\ \nu_p(d)=j}\nu_p(d)=\sum_{j=0}^{\nu_p(m_0)}\sum_{d\vert m_0\\ \nu_p(d)=j}j=\left(\sum_{j=0}^{\nu_p(m_0)}j\right)d(m_0p^{-\nu_p(m_0)})$$ $$\frac{1}{2}\nu_p(m_0)(1+\nu_p(m_0))d(m_0p^{-\nu_p(m_0)})=\frac{1}{2}\nu_p(m_0)d(m_0)=\frac{1}{2}\nu_p(M)|M|$$ So, if these are as bad as it gets, the previous conjectures follow. This also means that both inequalities (if true) are also sharp.
Note as well that both conjectures can be rephrased as generalization of Frankl's conjecture with multisets (or, equivalently, tuples of natural numbers) avoiding number-theoretic terminology. Specifically, if $\emptyset\neq E\subseteq \mathbb{N}^n$ is a finite max-closed set, i.e. $(x_j),(y_j)\in E\Rightarrow (\max\{x_j,y_j\})\in E$, then one can rephrase the previous conjectures as $$\array{\textbf{Conjecture 1. }\ \exists j\in n:|\{x\in E: x_j = 0\}|\leq \frac{1}{1+\max\limits_{x\in E} x_j}|E|}\\ \textbf{Conjecture 2. }\ \exists j\in n:\sum_{x\in E}x_j\geq \frac{1}{2}\left(\max\limits_{x\in E} x_j\right)|E|$$