When $\delta(\varepsilon)$ is written as you have above, it is merely a notational reminder that our choice of $\delta$ has to depend on the $\varepsilon$ we're given -- nothing more. In fact, $\delta$ also depends on $f$, $a$, and $L$. Writing $\delta(\varepsilon)$ does not mean that $\delta$ is a function to which we may plug in $\varepsilon$ to get our limit-satisfying $\delta$-value. In that vein, the also-common notation $\delta_\varepsilon$ could be argued to be better. However, we could create an actual function which acts in the spirit of the aforementioned $\delta(\varepsilon)$ and addresses your objection that we're "throwing out" other perfectly good values of $\delta$. This is most vivid if we restrict our attention to the following setup.
Let $A \subseteq \mathbb R$ be open and $f \colon A \longrightarrow \mathbb R$ have limit $L$ at $a$:
$$
\lim_{x \to a} f(x) = L.
$$
We may define
\begin{align}
\begin{split}
\delta_*(\varepsilon) &= \sup\{ \delta > 0 : a-\delta < x < a \implies |f(x) - L| < \varepsilon \}, \\
\delta^*(\varepsilon) &= \sup\{ \delta > 0 : a< x < a + \delta \implies |f(x) - L| < \varepsilon \},
\end{split}
\tag{1}
\end{align}
with the idea that $\delta_*(\varepsilon)$ tells you how far left of $a$ you can let $x$ go while keeping $|f(x) - L| < \varepsilon$, and $\delta^*(\varepsilon)$ tells you how far right of $a$ you can let $x$ go while keeping $|f(x) - L| < \varepsilon$. (We know that $\delta_*, \delta^* > 0$ exist because those sets on the RHS of $(1)$ are nonempty according to the limit definition.) Hence the largest open $x$-interval for which $|f(x) - L| < \varepsilon$ is $$
X(\varepsilon) = \big( a - \delta_*(\varepsilon), a + \delta^*(\varepsilon) \big).
$$
An issue here is that $X(\varepsilon)$ is not (necessarily) symmetric about $a$, so it doesn't (necessarily) correspond to $|x - a| < \delta$ for any $\delta$. To remedy this, we define $\hat \delta (\varepsilon) = \min\{\delta_*(\varepsilon), \delta^*(\varepsilon)\}$; then any $x$ in the interval
$$
X'(\varepsilon) = \big( a - \hat \delta(\varepsilon), a + \hat \delta (\varepsilon) \big)
$$
will satisfy $|f(x) - L| < \varepsilon$. Note that $X'(\varepsilon) = \{ x : |x - a| < \hat \delta(\varepsilon)\}$, and hence any $\delta$ in the interval $I(\varepsilon) = \big( 0, \hat \delta(\varepsilon) \big]$ satisfy the $\varepsilon$-$\delta$ definition of our limit. Moreover, $I(\varepsilon)$ is the largest set of values of $\delta$ that will work for a given $\varepsilon$. In other words:
$\delta$ satisfies the $\varepsilon$-$\delta$ definition $\iff \delta \in I(\varepsilon)$.
This answers your Question 1. A proof of this follows @grand_chat's answer. Note that $I(\varepsilon)$ depends on $a$, $f$, and $L$ implicitly.
One thing that may bother you is that $X(\varepsilon)$ may be much bigger than $X'(\varepsilon)$, so we're "throwing out perfectly good $x$'s". The $\varepsilon$-$\delta$ definition restricts $X(\varepsilon)$ to a symmetric interval ($X'(\varepsilon)$) about $a$. Does this help address your rigor question?
Of course satisfying the definition of a limit only requires us to find one such $\delta$. The reason that this is what you describe as the preferred method by professors etc. is the existence of complicated functions $f$ which make computing $I(\varepsilon)$ very difficult: it amounts to solving $f(x) = L \pm \varepsilon$ for $x$, which is inverting $f$. Since they don't need to find $I(\varepsilon)$, but just a single point in it, they opt for less work.
Your example of a "linear" function happens to be one in which the imprecise $\delta(\varepsilon)$ which people often write coincides with $\delta_*(\varepsilon) = \delta^*(\varepsilon) = \hat \delta (\varepsilon)$ in a quite canonical way, which may deceive people into believing some property of uniqueness for $\delta(\varepsilon)$.
Best Answer
You first need a suitable candidate/educated guess for what the limit ought to be. Then, only after that, you can use the precise definition to PROVE that your initial guess is indeed the case. Also, you can see that this is the best you can do simply from how the definition of limits is given:
Notice how the definition starts with "there exists $l\in \Bbb{R} \dots$" Just from the way it is phrased, it suggests that before even checking the $\epsilon,\delta$ criterion, you need to have a candidate value for the limit $l$. Nowhere does the definition tell you what $l$ is or how to go about guessing this (such "guess work" is something you pick up along the way as you learn more).
For example, if you had two functions $f$ and $g$, with $\lim\limits_{x\to a}f(x) = l_1$ and $\lim\limits_{x\to a}g(x) = l_2$, then if all you do is stare at the definition of limits, there's no way you can tell that $f+g$ also has a limit and that the limit equals $l_1+l_2$. The only natural guess would be that if $f+g$ had a limit, then it had better be $l_1+l_2$.
Then, once you have this guess, you then proceed to prove this using the precise $\epsilon,\delta$ definition (where the crux of the proof is the triangle inequality).