Primitive polynomials of finite fields there are two primitive polynomials which I can use to construct $GF(2^3)=GF(8)$:
$p_1(x) = x^3+x+1$
$p_2(x) = x^3+x^2+1$
$GF(8)$ created with $p_1(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3 = \alpha + 1$
$\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$
$\alpha^5 = \alpha^4 \cdot \alpha = (\alpha^2+\alpha) \cdot \alpha=\alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1$
$\alpha^6 = \alpha^5 \cdot \alpha=(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha+1+\alpha^2+\alpha=\alpha^2+1$
$GF(8)$ created with $p_2(x)$:
0
1
$\alpha$
$\alpha^2$
$\alpha^3=\alpha^2+1$
$\alpha^4=\alpha \cdot \alpha^3=\alpha \cdot (\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1$
$\alpha^5=\alpha \cdot \alpha^4=\alpha \cdot(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1$
$\alpha^6=\alpha \cdot (\alpha+1)=\alpha^2+\alpha$
So now let's say I want to add $\alpha^2 + \alpha^3$ in both fields. In field 1 I get $\alpha^2 + \alpha + 1$ and in field 2 I get $1$. Multiplication is the same in both fields ($\alpha^i \cdot \alpha^j = \alpha^{i+j\bmod(q-1)}$. So does it work so, that when some $GF(q)$ is constructed with different primitive polynomials then addition tables will vary and multiplication tables will be the same? Or maybe one of presented polynomials ($p_1(x), p_2(x)$) is not valid to construct field (altough both are primitive)?

The situation is not so different in a simpler context, the field of 5 elements, also known as the integers modulo 5. Whether $\alpha$ is $2$ or $3$, the field is $0,1,\alpha,\alpha^2,\alpha^3$, but whether $\alpha+\alpha+1=0$ depends on which $\alpha$ you choose. 
