Derive DFT of $x((n+1)/2)$
I would follow a signal block diagram based solution for this problem. 

First as suggested in the comments, it's very helpful to investigate a few values of $y[n]$ and $x[n]$ for some $n$ :

$$
\begin{align}
y[0] &amp;= 0 ~~~,~~~ y[1] = x[1] \\ 
y[2] &amp;= 0 ~~~,~~~ y[3] = x[2] \\
y[4] &amp;= 0 ~~~,~~~ y[5] = x[3] \\
y[6] &amp;= 0 ~~~,~~~ y[7] = x[4] \\
\end{align}
$$

with some experience, or by a direct investigation of this above sequence it can be seen that the system that produce $y[n]$ from $x[n]$ is the folowing :

$$ x[n] \longrightarrow \boxed{ \uparrow 2} \longrightarrow  w[n] \longrightarrow \boxed{ z^1} \longrightarrow y[n]$$

where the up arrow indicates an expansion by $2$ and the $z^1$ indicates a left shift (advance) by one sample.

Write down the DTFT relations between those signals $x[n]$,$w[n]$ and $y[n]$ :

$$
\begin{align}
W(\omega) &amp;= X(2\omega) \\ 
Y(\omega) &amp;= e^{j\omega} W(\omega) \\ \\
Y(\omega) &amp;= e^{j\omega} X(2\omega) \\
\end{align}
$$

And relate $2N$ point DFT $Y[k]$ of $y[n]$ to $N$ point DFT $X[k]$ of $x[n]$.

$$
\begin{align}
Y[k] &amp;= Y(\omega_k) = Y(\frac{2\pi}{2N} k) &amp;, ~~ k=0,1,...,2N-1\\
Y[k] &amp;= e^{j \frac{2\pi}{2N} k} X(\frac{2\pi}{N} k) &amp;, ~~ k = 0,1,...,2N-1\\
Y[k] &amp;= e^{j \frac{\pi}{N} k} X[k] &amp;, ~~ k = 0,1,...,2N-1
\end{align}
$$

It can be seen that the length $2N$ DFT sequence $Y[k]$ consists of two fold replica of length $N$ DFT sequence $X[k]$ weighted by $e^{j \frac{\pi}{N} k}$. Note that $X[k]$ is periodic in $k$ by $N$, hence repeats twice during $k = 0,1,...,2N-1$.