The question is very direct. What should $\alpha$ be for
$f(x)=\left\{ \begin{array} Xx^a\sin (x^{-b})&x \in (0,1]\\ 0&x= 0 \end{array} \right.$ to be $\alpha$-Hölder continuous?
It should be discussed by several situations. Just assume that $\alpha \le 1$. The situation $a<0$ is rather easy. Since $f$ must be continuous we have $b<a$. The Hölder condition on $x=0$ restrict $\alpha$ to be $\le a-b$. If $\alpha \le a-b$ then view $x^a \sin x^{-b}$ as $x^{\alpha \frac{a}{\alpha}} \sin x^{\alpha \frac{-b}{\alpha}}$ and use differential mean value theorem and the fact that $x_2^\alpha - x_1^\alpha \le (x_2 - x_1 )^\alpha$ for $x_2 > x_1$ it is easy to show that $f$ is $\alpha$-Hölder continuous.
For $a\ge 0$ and $b\le 0$ It is similar. But I find it hard to work out when $a > 0$ and $b > 0$.