In regards to finding a value for $\delta$ that satisfies the proof, we only really care about values of $x$ that are close to $4$, so setting $\delta < 1$ and $\delta < \frac{\epsilon}{61}$ is like saying the distance between $x$ and $4$ must be no more than $1$ or $\frac{\epsilon}{61}$.
By setting these restrictions on $\delta$, we can then make our lives easier in terms of solving this proof because the values acceptable for $x$ also become restricted, i.e if $\delta \lt 1 \rightarrow\lvert x - 4 \rvert \lt 1$ then its safe to make the claim that $3 \lt x \lt 5; x \not=4$ (because the difference between $x$ and $4$ cannot be greater than 1).
So, to answer question 1, "How can I get $\delta \lt \frac{\epsilon}{61}$ ?", it's the result we get when we restrict $\delta \lt 1$.
$$\begin{align}
\lvert ~f(x) - 64~\rvert &\lt \epsilon\\
\lvert ~x^3 - 64~\rvert &\lt \epsilon && \text{factor $x^3 - 64$} \\
\lvert ~(x - 4)(x^2 + 4x + 16)~| &\lt \epsilon && \text{isolate $(x - 4)$}\\
\lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{\lvert~x^2 + 4x + 16~\rvert}
\end{align}$$
Knowing that when $\delta \lt 1$, then $3 \lt x \lt 5; x \not=4$, so our goal is to apply this inequality to $x^2 + 4x + 16$ and see what we get.
$$\begin{align}
3 \lt x &\lt 5 && \text{let's find the min and max of $x^2$}\\
9 \lt x^2 &\lt 25 \\\\
3 \lt x &\lt 5 && \text{let's find the min and max of $4x$}\\
12 \lt 4x &\lt 20 \\\\
9 \lt x^2 &\lt 25 && \text{let's bring this all together now}\\
9 + 12 \lt x^2 + 4x &\lt 25 + 20 \\
21 \lt x^2 + 4x &\lt 45 && \text{all that's left to add is $16$}\\
21 + 16 \lt x^2 + 4x + 16 &\lt 45 + 16\\\\
37 \lt x^2 + 4x + 16 &\lt 61
\end{align}$$
What we just showed is that when $\delta \lt 1$, then $37 \lt x^2 + 4x + 16 \lt 61$ which in turn means that,
$$\begin{align}
\lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{\lvert~x^2 + 4x + 16~\rvert} \lt \frac{\epsilon}{61} \\\\
\lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{61} && \text{$\square$}
\end{align}$$
For your second question: "Why $x^3<\delta^3+12\cdot \delta^2 + 48\cdot \delta + 64 < \delta + 12\delta + 48\delta + 64 = 64 + \epsilon$ ?", if think about the value for $\delta$, it's always less than $1$ so when we when multiply two or more values that are less than $1$, we get a smaller number. This explains why,
$$\delta^3+12\delta^2 + 48\delta + 64 < \delta + 12\delta + 48\delta + 64$$
In regards to how $\delta + 12\delta + 48\delta + 64 = 64 + \epsilon$, we just simplify the right side of the equation to get
$$\begin{align}
\delta + 12\delta + 48\delta + 64 &= 61\delta + 64 && \text{sub. in $\frac{\epsilon}{61}$ for $\delta$}\\
61 \cdot \frac{\epsilon}{61} + 64 &= \epsilon + 64 && \text{$\square$}
\end{align}$$
Unfortunately, for your third question, I don't really know what your asking. However, what I can say about knowing when to set $\delta$ to either $1$ or $\frac{\epsilon}{61}$ is that it depends on the value for $\epsilon$; if $\epsilon \lt 61$ then set $\delta = \frac{\epsilon}{61}$, otherwise set $\delta = 1$.
Lastly, in terms of your proof having problems, these questions are not easy to begin with, so making a misstep in the beginning of your proof will very easily lead you astray. The good thing is that you tried to get the $\epsilon$ inequality to resemble the $\delta$ inequality which is the right thing to do, the problem is how you chose to do it.
Trying to isolate for $x$ and then manipulating the inequality to resemble the $\delta$ inequality isn't really an effective approach because things can get complicated very quickly (especially in this case). I would try and factor first and see where that takes me. That being said, when it comes to $\epsilon-\delta$ proofs, there is no silver bullet approach to finding the solution. You will need to practice them and with experience, you will start to have an idea of what to do.
As a side note, here's a reference to a tutorial on Limits I wrote a couple months ago for Mathbook.