Solution:

The transfer function of the filter is given by,

$\begin{aligned}\\ & H(z)= \sum_{n-0}^{N-1} h(n) z^{-n} \\\\ &= h(0)+h(1) z^{-1}+h(2) z^{-2}+h(3) z^{-3}+h(4) z^{-4}+h(5) z^{-5}+h(6) z^{-6} \\\\ &+h(7) z^{-7}+h(8) z^{-8} \\\\ & \alpha=\frac{N-1}{2}=\frac{9-1}{2}=4 .\\ \end{aligned}\\ $

$\begin{aligned}\\ H(z)= & z^{-4}\left[h(0) z^4+h(1) z^3+h(2) z^2+h(3) z^1+h(4) z^0+h(5) z^{-1}+h(6) z^{-2}\right. \\\\ & \left.+h(7) z^{-3}+h(8) z^{-4}\right]\\ \end{aligned}\\ $

Since $h(n)=h(N-1-n)$

$H(z)=z^{-4}\left[h(0)\left(z^4+z^{-4}\right)+h(1)\left(z^3+z^{-3}\right)+h(2)\left(z^2+z^{-2}\right)+h(3)\left(z+z^{-1}\right)+h(4)\right]\\ $

$\begin{aligned} H(\omega)= & e^{-j 4 \omega}\left[h(0)\left[e^{j 4 \omega}+e^{-j 4 \omega}\right]+h(1)\left[e^{j 3 \omega}+e^{-j 3 \omega}\right]+h(2)\left[e^{j 2 \omega}+e^{-j 2 \omega}\right]\right. \\\\ & \left.+h(3)\left[e^{j \omega}+e^{-j \omega}\right]+h(4)\right] \\\\ = & e^{-j 4 \omega}\left[h(4)+2 \sum_{n=0}^3 h(n) \cos (4-n) \omega\right] \\\\ = & e^{-j 4 \omega}|H(\omega)|\\ \end{aligned}\\ $

where $|H(\omega)|$ is the magnitude response and $\theta(\omega)=-5 \omega$ is the phase response. The phase delay $\tau_p$ and group delay $\tau_g$ are given by

$\tau_p=-\frac{\theta(\omega)}{\omega}=5 \text { and } \tau_g=\frac{d(\theta(\omega))}{d \omega}=-\frac{d(-5 \omega)}{d \omega}=5$