How to solve this even symmetry question?

I have my DSP final soon and I have been reviewing some past exams. Here is a question from one of these exams:

Let x[n] be a real valued finite duration signal in n $$\in [0,N]$$. Another signal $$x_1[n]$$ is defined as: $$x_1[n] = \begin{cases} x[n], & \text{if } n=0 \; or \; n=N ,\\ c x[n], & \text{if } n \in [1, \: N-1],\\ c x[2N-n] & \text{if } n \in [N+1,\: 2N-1] \end{cases}$$ where $$c$$ is a real valued constant and $$x_1[n]$$ is a finite duration signal $$\in [0, \: 2N-1]$$
a) Show that $$\mathrm{X}_1(k) = DFT_{2N}\{x_1[n]\}$$ is real valued$$.$$

b) Write $$\mathrm{X}_1(k) = DFT_{2N}\{x_1[n]\} = \sum_{n=0}^{N}x[n]\phi_k[n] \:, \; k = 0,...,2N-1$$ where $$\phi_k[n]$$'s have real values. Find those $$\phi_k [n]$$'s$$.$$

c) Show that $$\mathrm{X}_1(k) = \mathrm{X}_1(-k)_{\text{mod}\:2N}$$

Now I know that this question is testing the conjugate/even symmetric property of the DFT. I can handle c) but I am not sure how to approach parts a) and b). Here is my attempt for a) and solution for c):

a) The DFT is defined as follows: $$X[k] = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi nk/N}$$

In our case let $$2N=M$$ and plug in $$x_1[n]$$ into the DFT equation:

$$X_1[k] = \sum\limits_{n=0}^{M-1} x_1[n] e^{-j 2 \pi\frac nMk}$$ $$X_1[k] = x[0] + c\cdot \sum\limits_{n=1}^{N-1} x[n] e^{-j 2 \pi\frac nMk} + (x[N]\cdot(-1)^k) + c\cdot\underbrace{\sum\limits_{n=N+1}^{2N-1} x[2N-n] e^{-j 2 \pi\frac nMk}}_{G}$$ Let's work on $$G$$: $$G = \sum\limits_{n=N+1}^{2N-1} x[2N-n] e^{-j 2 \pi\frac nMk} \quad\quad \text{Replace n with n+N}$$ \begin{align} G &= \sum\limits_{n=1}^{N-1} x[N-n] e^{-j 2 \pi\frac {(n+N)}{M}k}\\ G &= (-1)^k\sum\limits_{n=1}^{N-1} x[N-n] e^{-j 2 \pi\frac nMk}\\ \end{align}

I don't know how to proceed from here. Perhaps it would be easier to prove that $$x_1[n]$$ is symmetric but I am not sure if that suffices.

Here is my solution for c) which I assume is wholly correct.

c) Just keep in mind that since $$x[n]$$ is real $$x_1[n]$$ is also real and also assuming that $$X_1(k)$$ is real as asked to prove in a). \begin{align} X_1(k) &= \sum_{n=0}^{M-1}x_1[n]e^{-j2\pi k\frac nM}\\ X_1^*(-k) &= \left(\sum_{n=0}^{M-1}x_1[n]e^{-j2\pi (-k)\frac nM}\right)^*\\ &= \sum_{n=0}^{M-1}\left(x_1[n]e^{-j2\pi (-k)\frac nM}\right)^*&\text{linearity}\\ &= \sum_{n=0}^{M-1}x_1[n]\left(e^{-j2\pi (-k)\frac nM}\right)^*&x_1[n]=x_1^*[n]\text{ since it's real}\\ &= \sum_{n=0}^{M-1}x_1[n]e^{-(-j2\pi (-k)\frac nM)}\\ &= \sum_{n=0}^{M-1}x_1[n]e^{-j2\pi k\frac nM}\\ &=X_1(k)_{\text{mod} \:2N} \end{align}

And help for a) and b) is much appreciated!

What you have shown for a) is actually what you should be doing for b). For a) you simply need to show that your given $$x_1[n]$$ is conjugate symmetric or even in this case.

Now for b) if we takeover from your $$G$$ then we can proceed as follows:

\begin{align*} G &= \sum\limits_{n=N+1}^{2N-1} x[2N-n] e^{-j 2 \pi\frac{n}{2N}k} \quad\quad \text{c.o.v. l=2N-n}\\ G &= \sum\limits_{l=1}^{N-1} x[l] e^{-j 2 \pi\frac {(2N-l)}{2N}k}\\ G &= \sum\limits_{l=1}^{N-1} x[l] e^{j 2 \pi\frac{l}{2N}k} \underbrace{e^{-j2\pi k}}_{(1)^k} \end{align*}

Now there is nothing stopping me from using $$l=n$$, it's a dummy variable anyways.

\begin{align*} G &= \sum\limits_{n=1}^{N-1} x[n] e^{j 2 \pi\frac{n}{2N}k} \end{align*}

You already have a sum running from $$n=1$$ to $$N-1$$, so combine the two under one summation:

\begin{align*} X_1[k] = x[0] + c\cdot \underbrace{\sum\limits_{n=1}^{N-1} x[n] e^{-j 2 \pi\frac nMk}}_{(i)} + (x[N]\cdot(-1)^k) + c\cdot\underbrace{\sum\limits_{n=1}^{N-1} x[n] e^{j 2 \pi\frac{n}{2N}k}}_{(ii)} \end{align*}

Combine $$(i)$$ and $$(ii)$$ as such:

\begin{align*} (i) + (ii) &= c\sum\limits_{n=1}^{N-1} x[n] (e^{j 2 \pi\frac{n}{2N}k} + e^{-j 2c \pi\frac{n}{2N}k})\\ (i) + (ii) &= 2\cdot \sum\limits_{n=1}^{N-1} x[n] \underbrace{\frac{(e^{j 2 \pi\frac{n}{2N}k} + e^{-j 2 \pi\frac{n}{2N}k})}{2}}_{\cos(\pi \frac nN k)} \end{align*}

So now we have shown that your $$\phi_k[n]$$'s are indeed real and we have also found all of them as such:

$$\phi_k[n] = \begin{cases} 1, & \text{if } n=0\\ 2c \cdot \cos(\pi \frac nN k), & \text{if } n \in [1, \: N-1]\\ (-1)^k, & \text{if } n=N\\ \end{cases}$$

for $$k \in \{0,1,...,2N-1\}$$.