= Solution
An extending <absolute value on a field> is non-Archimedean by Question 1(a). We prove <finite-dimensional non-Archimedean norm equivalence over a complete field>, using completeness rather than <local compactness>.
Fix a <basis> $e_1,\ldots,e_d$ of a finite-dimensional <vector space> over $K$, and let $\|\cdot\|$ be a <norm> homogeneous for the given <absolute value on a field> and satisfying the <ultrametric inequality>. Write $\|\sum a_i e_i\|_0=\max_i|a_i|$. The upper bound
$$
\|v\|\leq C\|v\|_0,\qquad C=\max_i\|e_i\|,
$$
is immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\ldots,e_{d-1}$ is complete for its restricted <norm> by the induction comparison with the coordinate <norm> and completeness of $K$. A complete <vector subspace> of a <metric space> is closed, so
$$
\delta=\inf_{w\in W}\|e_d-w\|>0.
$$
For $v=w+ae_d$ with $a\ne0$, scaling by $a$ gives $\|v\|\geq |a|\delta$; the same <coefficient> bound is trivial when $a=0$. Also
$$
\|w\|\leq\max(\|v\|,|a|\|e_d\|)
\leq\max(1,\delta^{-1}\|e_d\|)\|v\|.
$$
The induction bound controls every coordinate of $w$ by a constant times $\|w\|$. Together with the bound on $a$, this proves $\|v\|_0\leq C'\|v\|$.
Two extending <absolute values on a field> on $L$ are <norms> of this kind on its finite-dimensional <vector space> over $K$. Hence there is $B>0$ with $|x|_2\leq B|x|_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:
$$
|x|_2\leq B^{1/N}|x|_1.
$$
Letting $N\to\infty$ gives $|x|_2\leq|x|_1$. Interchanging the two <norms> gives equality. Thus
$$
\boxed{\text{There is at most one extending absolute value.}}
$$
The proof covers the trivial base <absolute value on a field> too. It proves uniqueness, without assuming an existence theorem or <compactness> of the coordinate sphere $\{v:\|v\|_0=1\}$.
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