DTFT of $(-1)^n \cdot \mathrm{sinc}()$
Considering @Stanley Pawlukiewicz's hint, you should take into account that $e^{~j \pi} = -1$, thus $(-1)^n = e^{~j \pi n}$, for any integer $n$.

Therefore, time-domain multiplication by $(-1)^n$ corresponds to a frequency-domain shift by $\pi$ radians in digital frequency.

Recall the Fourier transform's theorem: $\mathscr{F}\{x[n]~e^{j \omega_0 n}\} = X(\omega - \omega_0)$ where $\mathscr{F}\{\}$ is, in this case, the Discrete-Time Fourier Transform (DTFT).

Now, since

$$\mathscr{F}\left\{\frac{\sin (\pi n/2)}{\pi n}\right\} = \sum\limits_{k=-\infty}^{+\infty} \Pi\left(\frac{\omega + 2k\pi}{\pi}\right)$$

then, answering to your question:

$$\mathscr{F}\left\{(-1)^n \cdot \frac{\sin (\pi n/2)}{\pi n}\right\} = \sum\limits_{k=-\infty}^{+\infty}\Pi\left(\frac{\omega - \pi + 2k\pi}{\pi}\right)$$

which is a box function of amplitude $1$ and width $\pi$, centered on $\omega = \pi$.  The box function or rect function is 

$$\Pi(u) \triangleq \begin{cases}
0 &amp; \mbox{if } |u| &gt; \frac{1}{2} \\
\frac{1}{2} &amp; \mbox{if } |u| = \frac{1}{2} \\
1 &amp; \mbox{if } |u| &lt; \frac{1}{2}. \\
\end{cases}$$

Since the DTFT is periodic in $\omega$ with period $2\pi$, there is a box centered on $\omega = (2k-1)\pi$, for every integer $k$.