No. The important thing to know is that, if $K \subseteq L$ is a field extension and $v: K \to \mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $\mathbb{R}$. I'll work in $K = \mathbb{Q}(\sqrt{5})$ for the first question and in $K = \mathbb{Q}(\sqrt{3})$ for the second.

The ring of integers in $\mathbb{Q}[\sqrt{5}]$ is $\mathbb{Z}[\tau]$ where $\tau = \tfrac{1+\sqrt{5}}{2}$, with minimal polynomial $\tau^2=\tau+1$. Note that $\mathcal{O}_K/(2 \mathcal{O}_K)$ is the field $\mathbb{F}_4$ with four elements. Your first statement is true in $\mathbb{Q}$ only because $\mathbb{Z}/(2 \mathbb{Z})$ has two elements.

Specifically, both $1$ and $\tau$ are in $\mathcal{O}_K$ but not $2 \mathcal{O}_K$, so $v(1) = v(\tau) = 0$, but $1+\tau$ is also not in $2 \mathcal{O}_K$ so $v(1+\tau)=0$ as well.

Similarly, the ring of integers in $\mathbb{Q}(\sqrt{3})$ is $\mathbb{Z}[\sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+\sqrt{3})^2 (2-\sqrt{3})$ (note that $2-\sqrt{3}$ is a unit). We have $v(1) = v(\sqrt{3}) = 0$, but $v(1+\sqrt{3}^2) = 2$. In this case, the result is true in $\mathbb{Q}$ because $2$ is unramified.