For real

First, note that $f(8) = f(4 + 4) = [f(4)]^{2}$ and so $f(4) = 3^{\frac{1}{2}}$ as $f$ is positive valued.

Similarly, $f(2) = 3^{\frac{1}{4}}$ and $f(1) = 3^{\frac{1}{8}}.$

Now $f(nx) = [f(x)]^{n}$ , so $f(2015) = (3^{\frac{1}{8}})^{2015} = 3^{\frac{2015}{8}}.$

So $\log_{9}(f(2015)) = \dfrac{\log_{3}(f(2015))}{\log_{3}(9)} = \dfrac{2015}{16},$

and thus $a - b = 2015 - 16 = \boxed{1999}.$

f(x+y) = f(x)*f(y) is a fundamental property of exponential, so $f(x)=a^{x}$

in facts $a^{x+y} = a^{x}*a^{y}$ .

Now f(8) = 3 -> it means $a^{8} = 3$ and so $a = \sqrt[8]{3}$

than we have done, calcolate $log_9 ( \sqrt[8]{3})^{2015} = 2015 * log_9 \sqrt[8]{3} = \frac{2015}{16}$

and the result is 2015 - 16 = 1999

In fact, if there is a positive real-valued function $f(x+y)=f(x)+f(y)$ for all real $x,y$ . $f(x)$ must in a form of $f(x)=a^x$ .

Prove: $f(x+y)=f(x)+f(y)$

For any number $n$ ,

$f(n)\\=f(1+n-1)$

$=f(1)+f(n-1)\\=f(1)\cdot f(n-1)\\=f(1)\cdot [f(1)+f(n-2)]\\=f(1)\cdot f(1)\cdot f(n-2)$

$=[f(1)]^2\cdot f(n-2)$

$=[f(1)]^3\cdot f(n-3)$

$=[f(1)]^4\cdot f(n-4)$

$\cdots\\=[f(1)]^n$

$\rightarrow f(n)=[f(1)]^n$

By replacing $n$ and $f(1)$ with $x$ and $a$ , we get $\boxed{f(x)=a^x}(Q.E.D.)$