In a capacitor-charging circuit, let Q_1 denote the charge...

The charge on a capacitor during charging is given by $q(t) = Q_0(1 - e^{-t/RC})$.

The charging current is the rate of flow of charge, given by $i(t) = \dfrac{dq}{dt} = \dfrac{Q_0}{RC} e^{-t/RC}$.

Since the current $i(t)$ decreases exponentially with time, the rate at which charge is delivered to the capacitor continuously decreases.

For a fixed time interval, the charge delivered is the area under the $i(t)$ curve. Because $i(t)$ is a decreasing function, the charge delivered in the first $10 \text{ ms}$ interval is greater than the charge delivered in the subsequent $10 \text{ ms}$ interval. Thus, $Q_1 > Q_2$.

Similarly, to deposit a fixed amount of charge ($10 \mu\text{C}$), it takes less time initially when the current is high, and more time later when the current is lower. Therefore, the time $t_1$ required to deposit the first $10 \mu\text{C}$ is less than the time $t_2$ required to deposit the next $10 \mu\text{C}$. Thus, $t_1 < t_2$.