Transfer function of linear phase filter & its frequency domain representation
You finally have changed your $H(z)$ and added a qualifier "linear phase" , so is the new answer:

Now assuming that you have already determined $H(z)$ from the problem specifications, so the rest of obtaining DFT $H[k]$ can be approached in a number of ways one of which (perhaps longer than necessary) is the following:

Using the linearity and shift properties of the Z-Transform, we can show that that if a given signal $x[n]$ is of the form $$x[n] = a\delta[n] + b\delta[n-1] + c\delta[n-k]$$ then its Z-Transform is of the form: $$X(z) = a + b z^{-1} + c z^{-k}$$

Applying this observation, in the reverse direction, to $H(z)$ in your problem we get a prototype for $h[n]$ as: $$h[n]=k(\delta[n] - {5\over6} \delta[n-1] - {19\over3} \delta[n-2] - {5\over6} \delta[n-3] + \delta[n-4])$$

Also as you state that $h[0]=2$, yields $k=2$, hence the signal $h[n]$ is: $$h[n]= 2\delta[n] - {5\over3} \delta[n-1] - {38\over3} \delta[n-2] - {5\over3} \delta[n-1] + 2\delta[n-4]$$

Obtaining DFT $H[k]$ of this $h[n]$ is actually a very similar procedure to obtaining $H(z)$, you shall simply plug $e^{j{2\pi \over N} k}$ in place of $z$ where you simply get this:

$$H[k]= 2 - {5\over3} e^{-j{2\pi \over N} k} - {38\over3} e^{-j{4\pi \over N} k} -{5\over3} e^{-j{6\pi \over N} k} + 2e^{-j{8\pi \over N} k}$$

where $k$ ranges from $0$ to $N-1$ and plugging $N=5$ produces:
$$H[k]= 2 - {5\over3} e^{-j{2\pi \over 5} k} - {38\over3} e^{-j{4\pi \over 5} k} -{5\over3} e^{-j{6\pi \over 5} k} + 2e^{-j{8\pi \over 5} k}$$

Note that eventhough it is perfectly valid to state that, under suitable conditions, $H[k]=H(z)$ evaluated at $z=e^{j{2\pi\over N} k}$ , I guess it does not make an answer...