б) $$\frac{7}{45} + \frac{43}{45} \times \frac{7}{9} = \frac{7}{45} + \frac{301}{405} = \frac{7 \times 9}{45 \times 9} + \frac{301}{405} = \frac{63}{405} + \frac{301}{405} = \frac{364}{405}$$
Ответ: $$\frac{364}{405}$$