$\sin a \sin b$

$\sin (a+b)+\sin (a-b) =2\sin a\cos b=0.75+0.5=1.25$ .

$\sin (a+b)-\sin (a-b) =2\cos a\sin b=0.75-0.5=0.25$

So $\dfrac{\tan a}{\tan b}=\dfrac{1.25}{0.25}=\boxed 5$ .

$\begin{cases} \sin (a+b) = \sin a \cos b + \cos a \sin b = 0.75 = \frac 34 & ...(1) \\ \sin (a-b) = \sin a \cos b - \cos a \sin b = 0.5 = \frac 12 & ...(2) \end{cases}$

$\begin{cases} (1)+(2): & 2\sin a \cos b = \dfrac 54 & \implies \sin a \cos b = \dfrac 58 & ...(3) \\ (1)-(2): & 2\cos a \sin b = \dfrac 14 & \implies \cos a \sin b = \dfrac 18 & ...(4) \end{cases}$

$\begin{aligned} \implies \frac {(3)}{(4)}: \quad \frac {\sin a \cos b}{\cos a \sin b} & = \frac {\tan a}{\tan b} = \boxed 5 \end{aligned}$