${ 0.25, 0.125-j0.3018, 0, 0.125-j0.0518, 0 }.$

Determine the remaining three points

The given DFT is $\{0.25, 0.125 -j0.3018, 0, 0.125-j0.0150, 0\}$

Now, here

$X[0]= 0.25, \\ X[1]= 0.125 -j0.3018, \\ X[2]= 0, \\ X[3]= 0.125-j0.0150 \\ X[4]= 0$

By Symmetry Property of DFT,

$$X* [k ] = X[ N-k ]$$

Or

$$X* [ N-k ] = X[ k ]$$

Here,$ N = 8, \\ X* [k ] = X[8-k ] \\ X[ 5 ] = X*[ 8 - 5 ] = X* [3]\\ = 0.125+ j0.0150\\ X[ 6 ] = X*[ 8 - 6 ] = X* [2 ]\\ = 0\\ X[ 7 ] = X*[ 8 - 7 ] = X* [ 1 ]\\ = 0.125 + j0.3018$