Find the impulse response of an LTI system?
For the case a-) 
I assume you are trained enough in DSP to see that :
$$ \mathcal{ F } \{ \delta[n-d] \} = e^{-j \omega d } $$ for all integer $d$.

Hence given a frequency response of $H(\omega) = e^{-j3\omega }$ it's apparent that the corresponding impulse response is $$h_a[n] = \delta[n-3] $$

The problem in case b-) is in the fact that it suggests a non-integer amount of shift $\alpha$ of the unit impulse $\delta[n]$ so as to obtain $\delta[n-\alpha]$ which would then have a corresponding frequency response $H_b(\omega) = e^{ -j \omega \alpha }$. But this makes no sense in the domain of discrete-time sequences which cannot be shifted by non-integer amounts which might force you
to argue that the corresponding impulse response does not even exist.

The solution requires an investigation of the relation between continuous-time and discrete-time signals through sampling as the other answer outlines. Instead here I put a shorthand result.

First observe that for any integer $d$:
$$ \delta[n-d] = \frac{ \sin(\pi(n-d)) }{\pi(n-d) } = \text{sinc}(n-d)$$ is satisfied.

The righthand side is a sampled (and therefore discrete) sinc pulse whose continuous equivalent is $$\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$

Furthemore observe that the discrete-time sinc pulse $\sin(\pi n) /{ \pi n} $ is actually the impulse response of an ideal lowpass filter with a cutoff frequency of $\omega_c = \pi$ which is indeed an allpass filter with a frequency response of:
$$ H_{apf}(\omega) = 1 ~~~,~~~ \text{for all } ~~~|\omega| &lt; \pi $$   

As a consequence, the frequency response of a shifted sinc pulse
$ \frac{ \sin(\pi(n-d)) }{\pi(n-d) }$ will be: 

$$H_s(\omega) = e^{-j \omega d } $$ for any $d$, integer or real, according to the time-shift property of the Fourier transform.

Hence it can be concluded by the inverse Fourier transform relation of the last statement that the impulse response $h_b[n]$ associated with the frequency response $$H_b(\omega) = e^{-j \omega (3/2) }$$ is
$$h_b[n] =  \frac{ \sin(\pi(n-3/2)) }{\pi(n-3/2) } $$

Note that $h_b[n]$ cannot be obtained from $ \frac{ \sin(\pi n) }{\pi n}$ by just shifting it $3/2$ samples which is not possible. Therefore you should obtain the samples of the impulse response $h_b[n]$ by evaluating its formula for each n.