This is not an answer but just a set of comments:

A cousin of 1. is a yet unanswered question of J. D. Hamkins.

Notice that if an ordered field embeds into $\mathbf{No}$, then so does, as an ordered set, its ladder$^{\mathbf{[a]}}$. Since every linear order is the ladder of some ordered field, this means that a negative answer to 2. would answer the set-sized version of JDH's questions.

I assume by strictly larger ordered field you mean an ordered field which in addition does not embed in $\mathbf{No}$. Or do you mean that it should not inject into $\mathbf{No}$?
In the first case, notice that if $L$ is a linear order which does not embed into $\mathbf{No}$ then considering the ordered Hahn series group $G=\mathbb{R}[[\mathbf{No} \sqcup L]]$, the ordered Hahn series field $F=\mathbb{R}[[G]]$ is an example.

**[a]**: The ladder $L$ of an ordered field $F$ is obtained from $F^{\succ}:=\{x \in F: x > \mathbb{N}\}$ with the preorder $x\lll y$ iff $x^{\mathbb{N}}<y$.