On a certain day Mabel rode her bicycle for 45 minutes and then walked for 1 hours covering a total distance of 10.4km. the average cycling speed is three times the walking speed. Find correct to two significant figures the walking speed If Mabel increase the average cycling speed by 20% but maintain the walking speed for 1 hours find in minutes the new cycling time

Let $v_w$ be the walking speed in km/h.
Let $v_c$ be the cycling speed in km/h.
Cycling time $t_c = 45 \text{ minutes} = \frac{45}{60} \text{ hours} = \frac{3}{4} \text{ hours}$.
Walking time $t_w = 1 \text{ hour}$.
Total distance $D_{total} = 10.4 \text{ km}$.
Given that $v_c = 3v_w$.

Distance cycled $D_c = v_c \times t_c = (3v_w) \times \frac{3}{4} = \frac{9}{4} v_w$.
Distance walked $D_w = v_w \times t_w = v_w \times 1 = v_w$.

The total distance is the sum of the distances cycled and walked:

The walking speed, correct to two significant figures, is $\boxed{3.2 \text{ km/h}}$.

2. If Mabel increase the average cycling speed by 20% but maintain the walking speed for 1 hours find in minutes the new cycling time

Original walking speed $v_w = 3.2 \text{ km/h}$.
Original cycling speed $v_c = 3v_w = 3 \times 3.2 = 9.6 \text{ km/h}$.
New cycling speed $v_c' = v_c \times (1 + 0.20) = 9.6 \times 1.20 = 11.52 \text{ km/h}$.

Walking speed is maintained at $v_w = 3.2 \text{ km/h}$.
Walking time is maintained at $t_w = 1 \text{ hour}$.
Distance walked $D_w = v_w \times t_w = 3.2 \text{ km/h} \times 1 \text{ h} = 3.2 \text{ km}$.
Total distance remains $10.4 \text{ km}$.
Distance to be cycled $D_c' = D_{total} - D_w = 10.4 \text{ km} - 3.2 \text{ km} = 7.2 \text{ km}$.

New cycling time $t_c' = \frac{D_c'}{v_c'} = \frac{7.2 \text{ km}}{11.52 \text{ km/h}} = 0.625 \text{ hours}$.
Convert to minutes:

The new cycling time is $\boxed{37.5 \text{ minutes}}$.