In a math question I had such as the limit approaches $3$ in $x^4-x^2+1$ (help with formatting please I'm new), the answer was that $\delta$ equals $\left(1, \frac{\epsilon}{168}\right)$.
In another question, I have a limit approaching $5$ in $\frac{x+4}{x-4}$ and the answer equals $\left(\frac{1}{2}, \frac{\epsilon}{16}\right)$.
I can get the $\varepsilon$ answers pretty well but what do the $1$ and $\frac{1}{2}$ mean and how do you go about getting them?
Finding a suitable $\delta$ as a function of $\epsilon$ is by no means a straightforward process. Infinitely many suitable choices exist! It all depends on your strategy. One popular strategy that I like to use is to set $\delta = \min\{1, \epsilon/M\}$ for some suitable constant $M$. Note that the $1$ here is completely arbitrary; I could also have picked $2$, or $0.7$, or $1/2$, or $42$; I just prefer $1$ because it's a nice, small, round number. That seems to be the strategy used in your first example.
Given any $\epsilon > 0$, let $\delta = \min\{1, \epsilon/M\}$. [Note: The $M$ here is just a placeholder; we're going to go back and replace it with a concrete number once we figure out what it should be later.] Then observe that if $0 < |x - 3| < \delta$, then: \begin{align*} &|(x^4 - x^2 + 1) - 73| \\ &= |x^4 - x^2 - 72| \\ &= |x - 3||x^3 + 3x^2 + 8x + 24| \\ &< \frac{\epsilon}{M}|x^3 + 3x^2 + 8x + 24| &\text{since } |x - 3| < \delta \leq \frac{\epsilon}{M}\\ &= \frac{\epsilon}{M}|(x - 3)^3 + 12(x - 3)^2 + 53(x - 3) + 102| \\ &\leq \frac{\epsilon}{M}(|x - 3|^3 + 12|x - 3|^2 + 53|x - 3| + 102) &\text{by the triangle inequality} \\ &< \frac{\epsilon}{M}((1)^3 + 12(1)^2 + 53(1) + 102) &\text{since } |x - 3| < \delta \leq 1 \\ &= \frac{\epsilon}{M}(168) \\ &= \epsilon \end{align*} provided that $M = 168$. [Now we can go back in our proof and replace each $M$ with $168$.]
In the first one, if you compute $|f(x)-L|$ it is $|x-3)|\cdot q(x)$ where $q$ is a cubic, having (luckily) positive coefficients, and by restricting $x$ to be within $1$ of $3$ we can get to $q(x) \le 168$ because of $|x|\le 4.$ Thus we need to have both $|x-3|<1$ and $|x-3|<\varepsilon/168$, and the usual way to express two $x$ restrictions at once is by using the min function, so here you can say $\delta = \min(1,\varepsilon/168).$
The crux of your question seems to be, "how do you go about getting them" (the constants $1$ and $\frac 12$). The short answer is: the choice of constants $1$ or $\frac 12$ is a completely arbitrary choice in many cases; in other cases there are restrictions on what constant you can use, but the choice is still an arbitrary choice within those bounds. There is no formula to "derive" the $1$ or $\frac 12$ exactly, but that's OK, because you don't need a formula.
I've already explained the general idea in a previous answer, Why do we impose the constraint $\delta\le 1$ when proving the continuity of $x^3$ using the definition of continuity? That answer talked about a different function than $x^4 - x^2 + 1,$ but the main point is the same: the $1$ comes into play in $\delta = \min\left\{1, \frac{\epsilon}{168}\right\}$ because your $\delta$-$\epsilon$ assertion must give the desired result for all positive $\epsilon$, and a simpler assertion such as $\delta < \frac{\epsilon}{168}$ unfortunately would not be sufficient for large values of $\epsilon$ (specifically, in this example, any value of $\epsilon$ greater than $168$).
Your second problem is an example of "why $\delta < 1$ is not always helpful" as explained in my previous answer. The difficulty is that if $f(x) = \frac{x+4}{x-4},$ the function $f(x)$ is unbounded near $x = 4.$ If you are to find a limit of $f(x)$ at $x = 5,$ the condition $0 < \delta < 1$ tells you only that $|x - 5| < 1,$ that is, $4 < x < 6,$ and there are values of $x$ in that range (namely, values close to $4$) that will cause $| f(x) - f(5) |$ to be larger than whatever value of $\epsilon$ you may be given. But if you make a stricter condition, such as $0 < \delta < \frac 12,$ then you can avoid the bad behavior of $f(x)$ near $x = 4.$
