(1) Given that in triangle ABC, the sides opposite to angles A, B, and C are a, b, and c, respectively, and $$a=b\cos C+\frac{\sqrt{3}}{3}c\sin B$$.
Using the sine rule: $$2R\sin A=2R\sin B\cos C+\frac{\sqrt{3}}{3}2R\sin C\sin B$$,
Simplifying, we get: $\sin(B+C)=\sin B\cos C+\frac{\sqrt{3}}{3}\sin C\sin B$,
This implies: $\sin C\cos B=\frac{\sqrt{3}}{3}\sin C\sin B$,
Since $\sin C\neq 0$,
We get: $\tan B=\sqrt{3}$,
Given that $0<B<\pi$,
We find: $B=\frac{\pi}{3}$.

(2) Given that the area of triangle ABC is S=$$5\sqrt{3}$$,
This implies: $\frac{1}{2}ac\sin B=5\sqrt{3}$,
Solving for $ac$, we get: $ac=20$,
Since $a=5$,
We find: $c=4$.

Then, applying the cosine rule, we get:
$b^2=a^2+c^2-2ac\cos B$,
Substituting the known values, we get:
$b^2=25+16-20$,
Simplifying, we get:
$b^2=21$,
Solving for $b$, we get: $b=\boxed{\sqrt{21}}$.