Find h[n] using DTFT properties
While this is by your admission homework (and fairly basic), I'll bite. Recall the definition of the DTFT:

$$
X(\omega) = \sum_{n=0}^{\infty}x[n] e^{-j\omega n}
$$

And recall the definition of the frequency response $H(\omega)$:

$$
H(\omega) = \frac{Y(\omega)}{X(\omega)}
$$

where $x[n]$ is the input to the system and $y[n]$ is its output. Combine these two equations:

$$
\begin{eqnarray*}
H(\omega)X(\omega) &amp;=&amp; Y(\omega) \\
\frac{1 - a^4 e^{-j 4 \omega}}{1 - a^4 e^{-j \omega}}X(\omega) &amp;=&amp; Y(\omega) \\
(1 - a^4 e^{-j 4 \omega})X(\omega) &amp;=&amp; (1 - a^4 e^{-j \omega})Y(\omega) \\
X(\omega) - a^4 e^{-j 4 \omega} X(\omega) &amp;=&amp; Y(\omega) - a^4 e^{-j \omega} Y(\omega)
\end{eqnarray*}
$$

Now, perform the inverse DTFT on both sides of the equation. By definition, $X(\omega)$ and $x[n]$ are a transform pair; likewise for $Y(\omega)$ and $y[n]$. For the other two terms, recall the time-shifting property of the DTFT:

$$
x[n-k] \leftrightarrow e^{-jk\omega} X(\omega)
$$

which can be shown easily from the definition of the DFT. Using this property, the equation inverse transforms to the difference equation specification for the system:

$$
x[n] - a^4 x[n-4] = y[n] - a^4 y[n-1]
$$

$$
y[n] = x[n] - a^4 x[n-4] + a^4 y[n-1]
$$

This is the definition of a recursive filter, which are usually IIR; that is the case for this one. Finding the impulse response is easy; let $x[n] = \delta[n]$ and find that the system output is:

$$
y[n] = a^{4n}u[n] - a^{4(n-4)+4}u[n-4]
$$

$$
y[n] = a^{4n}u[n] - a^{4n-12}u[n-4]
$$



The above is plotted for $a=0.99$. It should be noted that the system is only stable for $|a|\le1$.