$H(e^{jω})=e^{j3ω} ;0≤|ω|≤\frac{π}{2}$ $H(e^{jω})=0 ;otherwise$

We know that

$H(k)=H(ω)_(ω=\frac{2πk}{N})$ ………(1)

$H(e^{jω} )=e^{-j3ω}$

Here ∝=3

$And ∝=\frac{(N-1)}{2}$

∴N=7

Now, By using equation (1)

$H(k)=e^{\frac{(-j6πk)}{7}} ;0≤ \frac{2πk}{7}≤\frac{π}{2}$

$=0 ; \frac{π}{2}≤\frac{2πk}{7}≤π$

$∴H(k)=e^{\frac{(-j6πk)}{7}} ;0≤k≤\frac{7π}{4π}$

$ =0 ; \frac{7π}{4π}≤k≤\frac{7π}{2π}$

$∴H(k)=e^{\frac{(-j6πk)}{7}} ;0≤k≤2$

$ =0 ;2≤k≤4$

Now , Here K is vary from 0 to 2

When k=0

$∴H(0)=1$

When k=1

$∴H(1)=e^{\frac{(-j6π)}{7}}$

When K=2

$∴H(2)=e^{\frac{(-j12π)}{7}}$

According to transfer function equation of Frequency Sampling Realization:

$H(z)=[\frac{(1-z^{-7})}{7}][\frac{H(0)}{(1-z^{-1}}+\frac{H(1)}{(1-e^{\frac{j2π}{7}} z^{-1}} +(\frac{H(2))}{(1-e^{\frac{j6π}{7}} z^{-1} )}]$