Convolving complex exponential with box function (discrete)
You can use the convolution theorem to convert the convolution to the product of Fourier transforms of the given signals.
$$h[n]*x[n]=\mathcal{F}^{-1}\{X(\omega)H(\omega)\}$$

$$H(\omega)=\mathcal{F}\{h[n]\}=\delta(\omega-\frac{\pi}{2})$$
$$X(\omega)=\mathcal{F}\{x[n]\}=e^{-j\frac{3 \omega}{2}}X_r(\omega)$$ where $$X_r(\omega)=\left\{\begin{matrix}
\frac{\sin(2\omega)}{\sin(\omega/2)} &amp;\omega\neq 0 \\ 
4 &amp; \omega=0
\end{matrix}\right.$$

However since $X(\omega)\delta(\omega-\frac{\pi}{2})=X(\frac{\pi}{2})\delta(\omega-\frac{\pi}{2})$ and $X(\frac{\pi}{2})=e^{-j\frac{3\pi}{4}}\frac{\sin(2\pi/2)}{\sin(\pi/4)}=0$, the inverse Fourier transform and therefore the covolution is zero. 

You can also assume $x[n]$ is the impulse response of an LTI system and $h[n]$ is an input signal. Since the complex exponential is eigenfunction of the system, the output whould be an scaled version of the input i.e. $Ah[n]$ where $A=X(\frac{\pi}{2})$ and $X$ is the FT of $x[n]$.