5.
$\frac{sin 3A+cos3A}{sin3A-cos 3A}=\frac{3(sin2A-1)}{cos2A}=\frac{3(2sinAcosA-sin^2A-cos^2A)}{cos^2A-sin^2A}=\frac{-3(cosA-sinA)^2}{cos^2A-sin^2A}=\frac{-3(cosA-sinA)}{cosA+sinA}$
$\Rightarrow \frac{(sin3A+cos3A)(cosA+sinA)}{(sin3A-cos3A)(cosA+sinA)}=-\frac{3(sin3A-cos3A)(cosA-sinA)}{(sin3A-cos3A)(cosA+sinA)}$
$\Rightarrow \frac{sin3AcosA+cos3AsinA+cos3AcosA+sin3AsinA}{(sin3A-cos3A)(cosA+sinA)}=-\frac{3(sin3AcosA+cos3AsinA-sin3AsinA-cos3AcosA)}{(sin3A-cos3A)(cosA+sinA)}$
$\Rightarrow \frac{sin(3A+A)+cos(3A-A)}{(sin3A-cos3A)(cosA+sinA)}=-\frac{3(sin(3A+A)-cos(3A-A))}{(sin3A-cos3A)(cosA+sinA)}$
$\Rightarrow \frac{4sin4A-2cos2A}{(sin3A-cos3A)(cosA+sinA)}=0$
$\Rightarrow \frac{2sin4A-cos2A}{(sin3A-cos3A)(cosA+sinA)}=0$
$\Rightarrow \frac{(4sin2A-1)(cos2A)}{(sin3A-cos3A)(cosA+sinA)}=0$
แต่ $-1<cos2A<0$
$\Rightarrow sin2A=\frac{1}{4} \Rightarrow cos2A=-\frac{\sqrt{15}}{4}$
$\Rightarrow \boxed{cos A = \sqrt{\frac{1-\frac{\sqrt{15}}{4}}{2}}=\sqrt{\frac{4-\sqrt{15}}{8}}\sim0.13}$ ($\because 0<cosA<\frac{\sqrt{2}}{2})$
$\sqrt{15}tan2A-tan2B=(\sqrt{15}-1)tan(A+B) \Leftrightarrow \sqrt{15}(tan2A-tan(A+B))-(tan2B-tan(A+B))=0$
$\Leftrightarrow \sqrt{15}(\frac{sin2A}{cos2A}-\frac{sin(A+B)}{cos(A+B)})-(\frac{sin2B}{cos2B}-\frac{sin(A+B)}{cos(A+B)})=0$
$\Leftrightarrow \sqrt{15}(\frac{sin(A-B)}{cos2A \ cos(A+B)})-\frac{sin(B-A)}{cos2B \ cos(A+B)}=0$
$\Leftrightarrow \frac{sin(A-B)}{cos(A+B)}(\frac{\sqrt{15}}{cos2A}+\frac{1}{cos2B})=0$
$\Leftrightarrow \frac{sin(A-B)}{cos(A+B)}(\frac{1}{cos2B}-4)=0$
$\Leftrightarrow \frac{sin(A-B)(1-4cos2B)}{cos(A+B)(cos 2B)}=0$
ได้ว่า $sin(A-B)=0$ หรือ $cos2B=\frac{1}{4}$
กรณี $sin(A-B)=0$ เนื่องจาก $-\frac{\pi}{2}<A-B<\frac{3\pi}{4} \Rightarrow A=B$
ดังนั้น $\boxed{sin B = sin A = \sqrt{\frac{{4+\sqrt{15}}}{8}}\sim0.99}$
กรณี $cos2B=\frac{1}{4} \Rightarrow \boxed{sinB=\sqrt{\frac{1-cos2B}{2}}=\sqrt{\frac{3}{8}}\sim 0.61}$