Think

Use the base switch rules, of which here I will use base $e$ . We will get $\frac{ln(25)}{ln(343)}\times\frac{ln(49)}{ln(125)} = \frac{2ln(5)}{3ln(7)}\times\frac{2ln(7)}{3ln(5)} = \frac{4}{9}$ Hence $a+b = 4+9 = 13$

$\log_{343} 25 = \frac{1}{3} \cdot \log_{7} 25 = \frac{2}{3} \cdot log_{7} 5$ ;

$\log_{125} 49 = \frac{1}{3} \cdot log_{5} 49 = \frac{2}{3} \cdot log_{5} 7$ ;

$\log _{343} 25 \times \log_{125} 49 = \frac{4}{9} \cdot \log_{7} 5 \times \log_{5} 7 = \frac{4}{9} \log_{7} 7 = \frac{4}{9}$ ... (4 + 9 =13)