Find Discrete Fourier transform given the inverse
METHOD-I 

Plain simple,

The discrete-time sequence you have provided is :

$$
f[n] = \begin{cases} 1 ~~~,~~~\text{ for } n = 0,1,2,14,15 \\ 0 ~~~,~~~ \text{ otherwise } \end{cases}
$$

which is also described as a sum of discrete-time unit impulses located for each $n=0,1,2,14,15$ as

$$f[n] = \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-14] + \delta[n-15] $$

Using the fact that, the N-point DFT of the unit-impulse located at $n=d$ is: $$ \boxed{ \delta[n-d] \longleftrightarrow X[k] = e^{-j \frac{2\pi}{N} k d} }$$

Then employ linearity property of DFT, to get the N-point DFT of $f[n]$ as:

$$
F[k] = e^{-j \frac{2\pi}{N} k 0} + e^{-j \frac{2\pi}{N} k 1} + e^{-j \frac{2\pi}{N} k 2} + e^{-j \frac{2\pi}{N} k 14} + e^{-j \frac{2\pi}{N} k 15}
$$    

which becomes with $N = 16$:
$$
F[k] = 1 + e^{-j \frac{\pi}{8} k} + e^{-j \frac{\pi}{4} k} + e^{-j \frac{\pi}{8} 14 k} + e^{-j \frac{\pi}{8} 15 k}
$$    

Now, not always possible but in this case you can merge those complex exponentials, after replacing the frequency of the the latter two according to the following.

$$ 
e^{-j \frac{\pi}{8} 14 k} = e^{-j \frac{\pi}{8} (16-2) k} = e^{-j 2\pi k} e^{j \frac{\pi}{4} k} = e^{j \frac{\pi}{4} k} 
$$

and similarly
$$
e^{-j \frac{\pi}{8} 15 k} = e^{-j \frac{\pi}{8} (16-1) k} = e^{-j 2\pi k} e^{j \frac{\pi}{8} k} = e^{j \frac{\pi}{8} k} 
$$

Then the DFT $F[k]$ becomes:
$$
F[k] = 1 + e^{-j \frac{\pi}{8} k} + e^{-j \frac{\pi}{4} k} +e^{j \frac{\pi}{8} k} + e^{j \frac{\pi}{4} k}
$$

Now combine those exponentials, with conjugate angles, into cosine terms according to $2 \cos(\omega n) = e^{j \omega n} + e^{-j \omega n}$ to conlcude:
$$
\boxed { F[k] = 1 + 2 \cos( \frac{\pi}{4} k) + 2 \cos( \frac{\pi}{8} k) ~~~,~~~\text{ for } k = 0,1,2,...,15} 
$$     

Which is a real and (circularly) even function of $k$, as the sequence $f[n]$ is (circularly) even-symmetric and real.

METHOD-II 

if you prefer a fancier approach, then you can also use this. By inspection, one can see that $f[n]$ is given by a circular left shift of another sequence $g[n]$ (i.e.,  $f[n] = g[n+d]$) where and $g[n]$ is

$$
g[n] = \begin{cases} 1 ~~~,~~~\text{ for } n = 0,1,2,3,4 \\ 0 ~~~,~~~ \text{ otherwise } \end{cases}
$$

and $d=2$. Then the N-point DFT of $f[n]$ and N-point DFT of $g[n]$ are related by (with $N=16$):

$$
F[k] = e^{j \frac{2\pi}{N} d k} G[k] ~~~,~~~\text{ for } k = 0,1,2,...,N-1 \\
F[k] = e^{j \frac{\pi}{4} k} G[k] ~~~,~~~\text{ for } k = 0,1,2,...,15 \\ 
$$

Ffurthermore, the N-point DFT of $g[n]$ is found as:
$$ 
G[k] = \sum_{n=0}^{4} 1 \cdot e^{-j \frac{2\pi}{16}nk} = \frac{ 1 - e^{-j \frac{5 \pi}{8} k} }{ 1 - e^{-j \frac{\pi}{8} k} }
$$

Finally the N-point DFT $F[k]$ of $f[n]$ becomes
$$ 
\boxed{
F[k] =e^{j \frac{\pi}{4} k} \left( \frac{ 1 - e^{-j \frac{5 \pi}{8} k} }{ 1 - e^{-j \frac{\pi}{8} k} }\right) ~~~,~~~\text{ for } k = 0,1,2,...,15
}
$$

Which should yield the same result.