The question is already answered very nicely by Steven Landsburg. The purpose of my answer is to add references and further results. Some of them are very general while others are very specific to **affine algebras**, the setting of your three questions.
A short and to-the-point answer could be: have a look to [2, VIII.2] (Projective Modules over Affine Algebras) and also to the end of [1, 11.5] (Suslin's theorems on affine algebras). I find these references (together with [3]) extremely useful. But I will elaborate, trying to address also some **practical** aspects.
Your first question asks about **practical** means to decide whether an element $e$ of the free module $A^n$, with $A$ an affine algebra, can be completed in a basis of $A^n$. The more general question where $A$ is replaced by an arbitrary unital ring $R$ (I will assume $R$ commutative for simplicity), has been investigated extensively in the context of the study of stably free modules, particularly in the resolution of Serre's problem on projective modules. Let us make this connection explicit.
An $R$-module $P$ is said to be *stably free of type $m$* ($0 \le m < \infty$) if $P \oplus R^m$ is free. The module $P$ is *stably free* if it is stably free of type $m$ for some $m$. We say that $e \in R^n$ is a *unimodular row* if the components of $e$ generated $R$ as an $R$-module (in other words, the row $e$ is a right-invertible matrix). We have the following
> **Proposition A [2, Corollary I.4.5].**
The following are equivalent:
> - Any finitely generated stably free $R$-module is free;
> - Any finitely generated stably free $R$-module of type $1$ is free;
> - Any unimodular row over $R$-module can be completed to an invertible matrix over $R$ (by adding a suitable number of rows).
Let us denote by $\text{Um}_n(R) \subset R^n$ the set of unimodular rows over $R$.The following proposition highlights the link between completable unimodular rows and the action of $\text{GL}_n(R)$ on $\text{Um}_n(R)$ by matrix right-multiplication.
> **Proposition B [1, Proposition 4.8].** The orbits of $\text{Um}_n(R)$ under the $\text{GL}_n(R)$-action are in $1$-$1$ correspondence with the isomorphism classes of $R$-modules $P$ for which $P \oplus R \simeq R^n$. Under this correspondence, the orbit of $(1, 0, \dots, 0)$ corresponds to the isomorphism classes of the free module $R^{n - 1}$. In particular $e \in \text{Um}_n(R)$ is completable if and only if $e \sim_{\text{GL}_n(R)} (1, 0, \dots, 0)$.
A ring which satisfies the equivalent properties of Proposition A is called a *Hermit ring* (T. Lam's terminology). So, if $A$ is a Hermit ring, then it suffices for $e$ to be unimodular in order to be part of a basis of $A^n$. Commutative semilocal rings, Dedekind rings and Bézout rings are Hermit rings [1, Examples I.4.7]. If $R/\text{rad}(R)$ is Hermit, then so is $R$. The latter reduction can be effective is some situations.
With Hermit rings we are of course asking for too much: we want unimodular rows of any size to be completable. Let us consider individual sizes.
First, there is a trivial, but effective observation: for any commutative unital ring $R$, any unimodular row of size $2$ is completable in a matrix of $\text{SL}_2(R)$. Focusing now on size-specific statements, we are led to the following definitions.
> **Definition [2, I.$(4.5)_d$].** Let $d$ be a non-negative integer.
A commutative unital ring $R$ is a *$d$-Hermit ring* if it satisfies any of the following equivalent properties.
> - Any finitely generated stably free $R$-module of rank $> d$ is free.
> - Any unimodular row over $R$-module of size $\ge d + 2$ can be completed to a (square) invertible matrix over $R$.
> - For $n \ge d + 2$, $\text{GL}_n(R)$ acts transitively on $\text{Um}_n(R)$.
> **Definition [1, 11.1.14].** The *general linear rank* $\text{glr}(R)$ of $R$ is the least integer $n \ge 1$ such that $\text{GL}_{n + 1}(R)$ acts transitively on $\text{Um}_{n + 1}(R)$.
A $d$-Hermit ring $R$ in the sense of T. Lam is thus a ring $R$ which satisfies $\text{glr}(R) \le d + 1$. As already observed by Steven Landsburg, a Noetherian ring $R$ of Krull dimension $d$ is a $d$-Hermit ring. This follows from Bass's Stable Range Theorem [2, Theorem II.7.3] [1, Theorem 11.3.7] [3, Section 1.1].
**I'll continue later.**
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[1] J. McConnell, J. Robson, "Noncommutative Noetherian Rings", 1987.
[2] T. Lam, "Serre's Problem on Projective Modules", 2006.
[3] C. Weibel, "The K-book: An introduction to algebraic K-theory", 2013.