$$g(x)h(x)=(x^2-4x+5)(x^{14}+5x^{13}+16x^{12}+40x^{11}+81x^{10}+125x^9+96x^8$$ $$\qquad -x^7+6x^5+25x^4+71x^3+160x^2+286x+355)$$ $$=x^{16}+x^{!5}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+240x^9+484x^8$$ $$+ x^7 + x^6 + x^5 + x^4 + x^3 + 11x^2 + 10x + 1775.$$ $g(2)=1$, $h(2)=378919$.

In fact, it is known that each polynomial with no nonnegative real roots has a multiple with only positive coefficients. Moreover, there exists a large variety of such polynomials, hence there is nothing special in such an example.

In fact

So, having obtained $h(x)$ such that the last few coefficients of $g(x)h(x)$ are large enough, you may change few its last coefficients to make $h(a)$ prime. That's how this concrete example was obtained: first, I have found one by one the coefficients of $h(x)$ so that the first coefficients of the product are ones and the next two are positive, then I have added the negative coefficient and repeated the procedure, and then I have changes the last coefficient to make $h(2)$ prime.

It might be helpful to notice the following fact. If $g(x)$ has a complex root with argument $\alpha$, then each its multiple $p(x)$ with nonnegative coefficients has the degree at least $\pi/\alpha$. To see this, just substitute that root into $p(x)$ and look at the sign of the imaginary part.

EDIT, I'm a bit lazy to construct an explicit example with the last coefficients $\pm1$, sorry; but here is the way.

Take $g(x)=x^2(x-2)^2+1$. In the same way you may find the polynomial $h(x)$ such that $g(x)h(x)$ has positive coefficients (and $h(x)$ does not) --- since $g(x)$ has no nonnegative real roots. Next, consider $g_1(x)=x^4g(1/x)$ and construct the corresponding polynomial $h_1(x)$, say of degree $n$ (we need $h_1$ to have $1$ as the leading coefficient; it is surely possible).

It remains to notice that $g(x)(x^{n+5}h(x)+x^nh_1(1/x))$ is almost the desired example; you just need to change the appropriate coefficients of $h$ and $h_1$ to make the corresponding value prime. I am almost sure thath this is possible...

Notice that one may even decrease the exponent $n+5$ (the only fact to check is that the second factor should have a negative coefficient, and that of $h(x)$ is somewhere in the middle).

1

$$g(x)h(x)=(x^2-4x+5)(x^{14}+5x^{13}+16x^{12}+40x^{11}+81x^{10}+125x^9+96x^8$$ $$\qquad -x^7+6x^5+25x^4+71x^3+160x^2+286x+355)$$ $$=x^{16}+x^{!5}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+240x^9+484x^8$$ $$+ x^7 + x^6 + x^5 + x^4 + x^3 + 11x^2 + 10x + 1775.$$ $g(2)=1$, $h(2)=378919$.

In fact, it is known that each polynomial with no nonnegative real roots has a multiple with only positive coefficients. Moreover, there exists a large variety of such polynomials, hence there is nothing special in such an example.

In fact, having obtained $h(x)$ such that the last few coefficients of $g(x)h(x)$ are large enough, you may change few its last coefficients to make $h(a)$ prime. That's how this concrete example was obtained: first, I have found one by one the coefficients of $h(x)$ so that the first coefficients of the product are ones and the next two are positive, then I have added the negative coefficient and repeated the procedure, and then I have changes the last coefficient to make $h(2)$ prime.

It might be helpful to notice the following fact. If $g(x)$ has a complex root with argument $\alpha$, then each its multiple $p(x)$ with nonnegative coefficients has the degree at least $\pi/\alpha$. To see this, just substitute that root into $p(x)$ and look at the sign of the imaginary part.