Water runs into an empty tank through a valve that is opened slowly and then left alone…

(a), (b) $r'(t) = 6e^{-t/2} - 3t\,e^{-t/2} = 6e^{-t/2}\left(1 - \dfrac t2\right)$. It is positive for $t < 2$ and negative for $t > 2$, so the rate is greatest at $t = 2\ \text{min}$. The greatest rate is $r(2) = 12e^{-1} = \frac{12}{e} \approx 4.41\ \text{L/min}$.

The antiderivative. With $u = t$ and $dv = e^{-t/2}\,dt$, so that $v = -2e^{-t/2}$: $\displaystyle\int 6t\,e^{-t/2}\,dt = 6\left(-2t\,e^{-t/2} + 2\int e^{-t/2}\,dt\right) = -12t\,e^{-t/2} - 24e^{-t/2} + C$.

(c) $\left[-12t\,e^{-t/2} - 24e^{-t/2}\right]_0^4 = (-48 - 24)e^{-2} + 24 = 24 - \frac{72}{e^{2}} \approx 24 - 9.744 = 14.26\ \text{L}$.

(d) The average rate is $\dfrac{14.256\ \text{L}}{4\ \text{min}} \approx 3.56\ \text{L/min}$.

(e) $\displaystyle\int_0^{b} 6t\,e^{-t/2}\,dt = 24 - (12b + 24)e^{-b/2}$, and $(12b + 24)e^{-b/2} \to 0$ as $b \to \infty$. So the tank receives $24.00\ \text{L}$ in all. More than half of it, about $14.3\ \text{L}$, comes in the first $4$ minutes.