Find $g$(3) if $g(x)g(y)=g(x)+g(y)+g(xy)-2$ and $g(2)=5$ 
If $g(x)$ is a polynomial function satisfying $g(x)g(y)=g(x)+g(y)+g(xy)-2$ for all real $x$ and $y$ and $g(2)=5$, then find $g$(3)

Putting $x=1,y=2$
$$
5g(1)=g(1)+5+5-2\implies \boxed{g(1)=2}
$$
And I think I can evaluate for $g(1/2)$ also.
Solution given in my reference is $g(3)=10$, but I think I am stuck with this.

The point is that $g$ is a polynomial.
Thus write $g(x) = a_n x^n + \dotsc + a_0$, and you have an equality:
$$\sum_{i, j = 0}^n a_i a_j x^i y^j = (\sum_{i = 0}^n a_i (x^iy^i + x^i + y^i)) - 2.$$
Since this is true for all $x, y$, it means that it is an equality as polynomials in $x, y$.
By comparing coefficients, we see that $a_i a_j$ must be zero whenever $i > j > 0$ (because there is no $x^i y^j$ term on the right).
Therefore $g$ is of the form $ax^n + b$, and our equation becomes:
$$a^2x^ny^n + abx^n + aby^n + b^2 = ax^ny^n + b + ax^n + b + ay^n + b - 2,$$
from which we deduce that $a = 0$ or $a = 1$.
If $a = 0$, then $b^2 = 3b - 2$, so that $b = 1$ or $b = 2$: this contradicts $g(2) = 5$. 
The only possibility is then $a = 1$. Comparing coefficients before $x^n$, we have $ab = a$, which tells us $b = 1$.
It turns out that any polynomial $g$ of the form $x^n + 1$ will satisfy that equation for all $x, y$.
From $g(2) = 5$ we then conclude that $n = 2$, hence $g(3) = 3^2 + 1 = 10$.
