For which $a>0$ series is convergent?

It is much simpler to use equivalents, which can be found with Taylor's formula:

• $2(1-\cos x)=x^2-\dfrac{x^4}{12}+o(x^4)$,
• $\sin(\sin x)=\sin\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\dfrac{x^3}6\Bigr)-\frac16\Bigl(x-\dfrac{x^3}6\Bigr)^{\!3}+o(x^3)=x-\dfrac{x^3}3+o(x^3)$, so $$x\sin(\sin x)=x^2-\frac{x^4}3+o(x^4).$$ Now, replacing $x$ with $\frac 1n$, we obtain $$2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)=\frac1{n^2}-\frac1{12n^4}-\frac1{n^2}+\frac1{3n^4}+o\biggl(\frac1{n^4}\biggr)=\frac1{4n^4}+o\biggl(\frac1{n^4}\biggr)$$ so an asymptotic equivalent for the general term of the series is $$\left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left(\sin\frac{1}{n}\right) \right)^a\sim_\infty\frac1{4^a n^{4a}}.$$

Knowing that series (with positive terms) which have asymptotically equivalent general terms both converge or both diverge, can you conclude?

The answer is indeed as you concluded $a > \frac{1}{4}$.

Note that (from the first line in your attempt)

$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$

for $n$ sufficiently large, and that is all you need to conclude

$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^{a} = \theta \left(\frac{1}{n^4} \right)^a = \frac{1}{n^{4a}},$$

as $\sum_{n=1}^{\infty} \theta \left(\frac{1}{n^4} \right)^a$ converges iff $a > \frac{1}{4}$.

To elaborate:

$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$

which implies for some positive constants $C_1, C_2$

$$ \frac{C_1}{n^4} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) \le \frac{C_2}{n^4} $$

which implies for positive $a$:

$$ \frac{C_1}{n^{4a}} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^a \le \frac{C_2}{n^4a} $$

However, $\sum_{n=1}^{\infty} \frac{C_1}{n^{4a}}$ diverges for all positive $a \leq \frac{1}{4}$ so if $a$ is positive then $a$ must satisfy $a > \frac{1}{4}$. On the other hand $\frac{C_2}{n^{4a}}$ converges for all $a > \frac{1}{4}$ so it suffices that $a > \frac{1}{4}$.

Can use a similar line of reasoning to show that $a$ cannot be nonnegative for the sum to converge.