The $n^3$ does "dominate" over the other terms in the cube root. That's a good observation.
However, observe that if $\frac{f(n)}{g(n)} \to 1$, it does not imply that $\frac{\sin f(n)}{\sin(g(n))} \to 1$ as $n \to \infty$ (the obvious counterexamples being say $2\pi n$ and $2\pi n + \frac{\pi}{2}$).
Therefore, domination of the kind you demonstrated is not the right kind of "domination" that you require to solve this answer. You need to dig deeper.
The right kind of domination is provided by the fact that if you want a "dominating" function $\sin (g(n))$ to capture the behaviour of the function $\sin(f(n))$, then what you want precisely is that $\frac{\sin f(n)}{\sin g(n)} \to 1$.
What the counterexample tells you is that if $f,g$ are two functions for which you want the relation $\frac{\sin f(n)}{\sin g(n)} \to 1$ to hold, then it's not even sufficient to have $f(n)-g(n)$ bounded as $n \to \infty$. It must go to zero : and at a certain rate.
What rate? The following lemma is the answer.
Lemma : Suppose that $a_n, n \geq 1$ is an increasing sequence of real numbers , $a_n \to \infty$. Suppose that $f,g$ are functions taking positive values such that $a_n$ lie in the domain of both $f$ and $g$. If $$
\lim_{n \to \infty} \frac{|f(a_n)-g(a_n)|}{\sin g(a_n)} = 0
$$
then $$
\lim \frac{ \sin f(a_n)}{\sin g(a_n)} =1
$$
In particular, suppose that $\lim_{n \to \infty} \sin g(a_n) \neq 0$ and $\lim_{n \to \infty} \sin f(a_n) - \sin g(a_n) = 0$. Then, the above conclusion holds.
Proof : By the mean value theorem for the $\sin$ function applied at the points $f(a_n)$ and $g(a_n)$ (and using the fact that $|\cos x| \leq 1$ for all $x$), $\left|\sin f(a_n) - \sin g(a_n)\right| \leq |f(a_n)-g(a_n)|$, so that $$
\left|\frac{\sin f(a_n)}{\sin g(a_n)} - 1\right| \leq \frac{|f(a_n)-g(a_n)|}{g(n)} \leq \frac{|f(n)-g(n)|}{g(n)}
$$
whence we are done by the squeeze theorem. $\blacksquare$
So given a "complicated" function $f$, the "dominating" term $g$ will have to be much, much better than just satisfying $\frac{f(n)}{g(n)} \to 1$. The above lemma expresses this as well.
Now to solve the first problem, where we have $f(a_n) = a_n\pi\sqrt[3]{a_n^3+3a_n^2+4a_n-5}$ and $a_n = n$ (we are along the integers).
To find $g$, we use the fact that $n^3+3n^2+4n-5$ is very close to $n^3+3n^2+3n+1 = (n+1)^3$. Therefore, we write $$
n^3+3n^2+4n-5 = (n+1)^3 \sqrt[3]{1+\frac{n-6}{(n+1)^3}}
$$
whence $$
f(n) = n(n+1)\pi \sqrt[3]{1+\frac{n-6}{(n+1)^3}}
$$
Now, we are tempted to take $g(n) = n(n+1)\pi$ : however, observe that $n(n+1)\pi$ is always an even multiple of $\pi$, therefore $\sin(g(n)) = 0$ for all $n$. There's no way that the limit in the lemma can exist!
Therefore, we need to go one step further. To do this, we require the Taylor expansion of $\sqrt{3}{1+\frac{n-6}{(n+1)^3}}$ around the point $1$. If $y_n = \frac{n-6}{(n+1)^3}$ then the Taylor expansion to the first order gives $$
\sqrt[3]{1+y_n} = 1+\frac{y_n}{3} + o(y_n)
$$
where $o(y_n)$ is a function of $n$ which , when divided by $y_n$, goes to $0$ as $n \to \infty$.
Putting this into $f(n)$, $$
f(n) = n(n+1)\pi + \frac{n(n+1)\pi y_n}{3} + o(n(n+1)\pi y_n)
$$
Finally putting back the value of $y_n$ for the middle and last terms, $$
f(n) = n(n+1)\pi +\frac{\pi}{3}\frac{n(n+1)(n-6)}{(n+1)^3}+o\left(\frac{n(n+1)(n-6)\pi}{(n+1)^3}\right)
$$
Observe each term now : if we let $$
g(n) = n(n+1)\pi +\frac{\pi}{3}\frac{n(n+1)(n-6)}{(n+1)^3}
$$
then it's easy to see that $\frac{n(n+1)(n-6)}{(n+1)^3} \to 1$ as $n \to \infty$. Therefore, by the periodicity of sine, $\sin g(n) \to \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ as $n \to \infty$. On the other hand, observe that $$
f(n)-g(n) = o\left(\frac{n(n+1)(n-6)\pi}{(n+1)^3}\right)
$$
The right hand side , once you see the powers of $n$ on the top and bottom, is just $o(1)$. Therefore, $|f(n)-g(n)| \to 0$ as $n \to \infty$.
Thus, the lemma can be applied and $\frac{\sin f(n)}{\sin g(n)} \to 1$. However, using the produce rule, this is the same as $\sin f(n) \to \frac{\sqrt 3}{2}$, as desired.
What about the second sequence? Here we have $a_n = n+0.6$.
Notice something very important : here, when we use algebra similar to the previous section and write $$
f(a_n) = a_n(a_n+1)\pi\sqrt[3]{1+\frac{a_n-6}{(a_n+1)^3}}
$$
and think of taking $g(a_n) = a_n(a_n+1)\pi$, then the limit $\sin g(a_n)$ doesn't even exist. Hence, this $g$ won't work.
However, even if we Taylor expand $g$, it turns out that we can't get rid of the bad term $a_n(a_n+1)\pi$, which is creating all the periodicity.
Therefore, we are inclined to think that , despite the apparent evidence to the opposite, $\lim_{n \to \infty} f(n+0.6)$ doesn't exist. That is, not only is the lemma not applicable, the end result isn't true either.
To prove this rigorously, we should demonstrate explicitly two different subsequences of $n+0.6$ for which $f$ converges to different values.
Let's see how we can do that. Clearly, we may still write using the same Taylor expansion as in the previous proof that $$
f(n) = g(n)+ o\left(\frac{n(n+1)(n-6)\pi}{(n+1)^3}\right)
$$
and simply replace $n$ with $n+0.6$ to obtain $$
f(n+0.6) = g(n+0.6)+o\left(\frac{(n+0.6)(n+1.6)(n-5.4)\pi}{(n+1.6)^3}\right)
$$
However, $g(n+0.6) = (n+0.6)(n+1.6)\pi + \frac{\pi}{3}\frac{(n+0.6)(n+1.6)(n-5.4)\pi}{(n+1.6)^3}$. The second term is small, so if we can get the first term to behave badly, we can get the whole thing to behave badly.
To do that, let's observe that $$
g(n+0.6) = n(n+1)\pi + \left(1.2n + 0.96\pi\right)\pi + \frac{\pi}{3}\frac{(n+0.6)(n+1.6)(n-5.4)\pi}{(n+1.6)^3}
$$
We can use two different subsequences now : let $a_n = 5,10,15,20,25,\ldots$ and let $b_n=1,6,11,16,\ldots$. It can be verified that $$
\sin g(a_n) \approx \sin \left(0.96\pi\right)
$$
while $$
\sin g(b_n) \approx \sin \left(0.16 \pi\right)
$$
In particular, taking each of these $a_n$ and $b_n$, the limit $\lim_{n \to \infty} g(n+0.6)$ does not exist. (Interestingly enough, observe that the noted subsequential limits do exist, so the lemma in the first section can be used to prove limit theorems for $f(a_n),f(b_n)$.)
However, because of the mean value theorem for $\sin$, $\sin f(n+0.6) - \sin g(n+0.6) \to 0$ as $n \to \infty$. This shows that $\sin f(n+0.6)$ does not exist.
Using a similar argument, one can easily show
Suppose that $r$ is any non-zero rational number, or any number such that $\frac{r}{\pi}$ is irrational. Then, $\lim_{n \to \infty} f(n+r)$ does not exist, where $f(x) = x \pi \sqrt[3]{x^3+3x^2+4x-5}$
The lemma can be used in a lot more general situation (and can be generalized to more functions than the sine) than has been done here, of course.
