8、解：$\because a^{3}+b^{3}+7ab=9$
$\therefore (a+b)(a^{2}-ab+b^{2})+7ab=9$
$\therefore (a+b)\left[ (a+b)^{2}-3ab \right]+7ab=9$
设 $t=a+b$
$\therefore t^{3}-3tab+7ab=9$
$\therefore \dfrac{t^{3}-9}{3t-7}=ab\leq\dfrac{t^{2}}{4}$
$\therefore \dfrac{t^{3}+7t^{2}-36}{4(3t-7)}\leq 0$
$\therefore \dfrac{(t^{3}-8)+(7t^{2}-28)}{4(3t-7)}\leq 0$
$\therefore \dfrac{(t-2)(t^{2}+9t+18)}{4(3t-7)}\leq 0$
$\therefore 2\leq t<\dfrac{7}{3}$
9、解：$\because 2\sin A+\sin B=2\sin C$
$\therefore \sin B=2(\sin C-\sin A)$
$\therefore \sin(A+C)=2(\sin C-\sin A)$
$\because \sin\dfrac{C-A}{2}=\sin\dfrac{C}{2}\cos\dfrac{A}{2}-\cos\dfrac{C}{2}\sin\dfrac{A}{2}$
$\cos\dfrac{C+A}{2}=\cos\dfrac{C}{2}\cos\dfrac{A}{2}-\sin\dfrac{C}{2}\sin\dfrac{A}{2}$
$\therefore \sin\dfrac{C-A}{2}\cdot\cos\dfrac{C+A}{2}=\dfrac{1}{2}\sin C-\dfrac{1}{2}\sin A$
$\therefore 2\sin\dfrac{C+A}{2}\cos\dfrac{C+A}{2}=4\sin\dfrac{C-A}{2}\cdot\cos\dfrac{C+A}{2}$
$\therefore 2\sin\dfrac{C+A}{2}=4\sin\dfrac{C-A}{2}$
$\therefore \tan\dfrac{C}{2}=3\tan\dfrac{A}{2}$
设 $m=\tan\dfrac{A}{2}$，则 $\tan\dfrac{C}{2}=3m$
$\therefore \dfrac{5}{\sin A}+\dfrac{9}{\sin C}=\dfrac{5}{\dfrac{2m}{m^{2}+1}}+\dfrac{9}{\dfrac{6m}{9m^{2}+1}}=\dfrac{16m^{2}+4}{m}=16m+\dfrac{4}{m}\geq 16$
当且仅当 $m=\tan\dfrac{A}{2}=\dfrac{1}{2}$ 时等号成立。