Calculate, for the entire journey, the average speed of the shuttle.

The total distance is $300 \text{ km}$.
The distance for the first half is $D_1 = \frac{300 \text{ km}}{2} = 150 \text{ km}$.
The speed for the first half is $v_1 = 60 \text{ km/h}$.
The time taken for the first half is $T_1 = \frac{D_1}{v_1}$.

$$T_1 = \frac{150 \text{ km}}{60 \text{ km/h}} = 2.5 \text{ hours}$$

The distance for the second half is $D_2 = \frac{300 \text{ km}}{2} = 150 \text{ km}$.
The speed for the second half is $v_2 = 90 \text{ km/h}$.
The time taken for the second half is $T_2 = \frac{D_2}{v_2}$.

$$T_2 = \frac{150 \text{ km}}{90 \text{ km/h}} = \frac{15}{9} \text{ hours} = \frac{5}{3} \text{ hours}$$

The total time $T_{total}$ is the sum of the times for the first and second halves.

$$T_{total} = T_1 + T_2 = 2.5 \text{ h} + \frac{5}{3} \text{ h} = \frac{5}{2} \text{ h} + \frac{5}{3} \text{ h}$$

To add these fractions, find a common denominator, which is 6.

$$T_{total} = \frac{5 \times 3}{2 \times 3} \text{ h} + \frac{5 \times 2}{3 \times 2} \text{ h} = \frac{15}{6} \text{ h} + \frac{10}{6} \text{ h} = \frac{25}{6} \text{ hours}$$

The average speed $v_{avg}$ is the total distance divided by the total time.

$$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{300 \text{ km}}{\frac{25}{6} \text{ h}}$$ $$v_{avg} = 300 \text{ km} \times \frac{6}{25 \text{ h}} = \frac{1800}{25} \text{ km/h} = 72 \text{ km/h}$$

The average speed for the entire journey is $\boxed{\text{72 km/h}}$.

The distance for the first half remains $D_1 = 150 \text{ km}$.
The new speed for the first half is $v'_1 = 50 \text{ km/h}$.
The new time taken for the first half is $T'_1 = \frac{D_1}{v'_1}$.

$$T'_1 = \frac{150 \text{ km}}{50 \text{ km/h}} = 3 \text{ hours}$$

The time for the second half remains $T_2 = \frac{5}{3} \text{ hours}$ (from part a).
The new total time $T'_{total}$ is the sum of the new time for the first half and the time for the second half.

$$T'_{total} = T'_1 + T_2 = 3 \text{ h} + \frac{5}{3} \text{ h}$$ $$T'_{total} = \frac{9}{3} \text{ h} + \frac{5}{3} \text{ h} = \frac{14}{3} \text{ hours}$$

The original total time was $T_{total} = \frac{25}{6} \text{ hours}$ (from part a).
The change in overall time is $\Delta T = T'_{total} - T_{total}$.

$$\Delta T = \frac{14}{3} \text{ h} - \frac{25}{6} \text{ h}$$

To subtract these fractions, find a common denominator, which is 6.

$$\Delta T = \frac{14 \times 2}{3 \times 2} \text{ h} - \frac{25}{6} \text{ h} = \frac{28}{6} \text{ h} - \frac{25}{6} \text{ h} = \frac{3}{6} \text{ h} = \frac{1}{2} \text{ hour}$$

To express this in minutes:

$$\frac{1}{2} \text{ h} \times 60 \text{ min/h} = 30 \text{ minutes}$$

The overall time increases by $\boxed{\text{30 minutes}}$.