Multiplying in GF(128) I know that in GF(128) $a + b = a \oplus b$.
I have an multiplication table for GF(128). In this table $7\cdot 5 = 27$. How can I create table like this?
 A: In general the multiplication depends on what defining polynomial you use. For the example that you give that is not immediately needed as the factors have such a low degree. My educated guess is that whoever gave that table is internally using a monomial presentation, and converts the sequence of coefficients to integers. Thus $7=111_2$ truly means the coset of the polynomial 
$$1\cdot x^2+1\cdot x+1=x^2+x+1$$ 
(modulo some ideal $I$ of the ring $\Bbb{Z}_2[x]$). Similarly 
$$5=101_2=1\cdot x^2+0\cdot x+1=x^2+1.$$
Therefore their product is the coset of
$$
\begin{aligned}
"7"\cdot"5"&=(x^2+1)(x^2+x+1)=(x^4+x^3+x^2)+(x^2+x+1)\\
&=x^4+x^3+2x^2+x+1\\
&=x^4+x^3+x+1,
\end{aligned}
$$
because the arithmetic of the coefficiente of polynomials is done modulo two. Note that
$x^4+x^3+x+1$ corresponds to $11011_2=27$.
You get the field $GF(128)$ if you do all the arithmetic as polynomials of degree at most six modulo two and reduce the high degree ($\ge7$) terms using the defining polynomial of degree seven (that seven comes from $128=2^7$). I usually use $x^7+x^3+1$ when I need $GF(128)$. This means that (for example)
$$
x^4\cdot x^3=x^7=x^7+(x^7+x^3+1)=2x^7+x^3+1=x^3+1,
$$
so $"16"\cdot"8"=1001_2="9"$, and
$$
x^4\cdot x^5=x^9=x^9+x^2(x^7+x^3+1)=2x^9+x^5+x^2=x^5+x^2,
$$
so $"16"\cdot"32"=x^5+x^2=100100_2="36"$.
In other words you declare that $x^7=x^3+1$, use that and all its consequences such as
$x^8=x^4+x$, $x^9=x^5+x^2$ et cetera.
