Inverse DTFT of $H_1(\Omega)=\begin{cases} 10,& \frac{\pi}{3} \leq |\Omega| < \pi\\ 0,& 0 \leq |\Omega| < \frac{\pi}{3}\\ \end{cases}$
Let's use $\omega$ instead of $\Omega$ for the definition of the inverse DTFT and state the following:

$$h[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} H(\omega) e^{j \omega n} d\omega $$ Given that $H(\omega) = 1 $ , for $\pi/3 &lt; |\omega| &lt;\pi $ and zero otherwise, the integral becomes:

$$\begin{align}
h[n] &amp;= \frac{1}{2\pi} \int_{-\pi}^{-\pi/3} e^{j \omega n} d\omega + \frac{1}{2\pi} \int_{\pi/3}^{\pi} e^{j \omega n} d\omega \\ 
&amp;= \frac{1}{2\pi} \left(  \frac{ e^{j \omega n} }{j n} |_{-\pi}^{-\pi/3} + \frac{ e^{j \omega n} }{j n} |_{\pi}^{\pi/3}  \right)\\  
&amp;= \frac{1}{2\pi} \left(  \frac{ e^{-j (\pi/3) n} -e^{-j\pi n}  }{j n} + \frac{ e^{j \pi n} - e^{j \pi/3 n} }{j n}  \right)\\ 
&amp;= \frac{1}{n\pi} \left(  \frac{ e^{-j (\pi/3) n}- e^{j (\pi/3) n} }{2j} + \frac{ e^{j \pi n} - e^{-j\pi n} }{2j}  \right)\\  
&amp;= \frac{1}{n\pi} \left(  -\sin(\frac{\pi}{3}n) + \sin(\pi n) \right)\\
&amp;= \frac{\sin(\pi n) }{n\pi} - \frac{ \sin(\frac{\pi}{3}n)}{n\pi}  \\
\end{align}
$$

at this point recognize  that 
$$\frac{\sin(\pi n) }{n\pi} = \begin{cases} 1 &amp;,\text{ for } n=0 \\0 &amp;, \text{ for } n \neq 0 \\ \end{cases} = \delta[n]$$

Hence the I-DFT becomes:
$$ \boxed{ h[n] =  \delta[n] - \frac{\sin( \frac{\pi}{3}n) }{\pi n} }$$

Note that your $H(\omega)$ had a gain of $10$ instead of $1$ so multiply this with $10$.