The sum of $n$ consecutive terms in an arithmetic sequence must be in the form of $An^2 + Bn$ where $A$ and $B$ are two constants. Let $S_n = A_1 n^2 + B_1$ and $T_n=A_2n^2 + B_2$, when $A_1, B_1, A_2,$ and $B_2$ are to be determined constant coefficients. Therefore $$\frac{S_n}{T_n}=\frac{A_1 n^2 + B_1n}{A_2 n^2 + B_2n}=\frac{2n}{3n+7}$$ It follows that $B_1 = 0$ and $A_1:A_2:B_2=2:3:7$. Let $A_1 = 2k, A_2 = 3k,$ and $B_2=7k$ where $k$ is a constant. $$S_n = 2kn^2 \implies a_8 = S_8 - S-7 = 128k - 98k=30k$$ $$T_n = 3kn^2 + 7kn \implies b_6=T_6-T_5 = (108k + 42k)-(75k+35k)=40$$ $$\therefore\quad \frac{a_8}{b_6}=\frac{30k}{40k}=\boxed{\frac{3}{4}}$$