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?

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.
