Hidden zero in system equation $H(z)$
If the problem description is correct, i.e., if $h[n]$ is zero for $n&lt;0$ and $n&gt;7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $z=-1$, in the second case it must have a zero at $z=1$. Refer to this answer for more details on the $4$ types of linear phase FIR filters.

So if we assume even symmetry (type II), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=-1$.

If we assume odd symmetry (type IV), the zeros are $z_0=0.4e^{j\pi/3}$, $z_1=z_0^*$, $z_2=1/z_0$, $z_3=1/z_0^*$, $z_4=3$, $z_5=1/z_4$, $z_6=1$.

The corresponding impulse responses are

$$h_{II}=[1,   -5.23,   11.84,  -12.42,  -12.42,   11.84,   -5.23,    1]$$

$$h_{IV}=[1,   -7.23,   24.31,  -48.57,   48.57,  -24.31,    7.23,   -1]$$


EDIT:
The answer above is correct (as far as I can tell) for the given problem description. Now that I've checked the original source (Schaum's Outlines of DSP), which includes the solution, I believe that there are the following possibilities: either there's a typo in the problem description and they actually meant that $h[n]$ is zero for $n&lt;0$ and $n\ge 7$ (note the "greater or equal" sign). In this case we really have a $6^{th}$ order FIR filter and the given solution in Schaum's Outline is correct. The other option would be that they don't know what they're talking about (and I hope and believe that this is not the case).