Your answers for (a) and (c) are correct.
(b) Salt flows into the tank at a rate of
[tex]\left(0.025 \dfrac{\rm kg}{\rm L}\right) \left(5 \dfrac{\rm L}{\rm min}\right) = 0.125 \dfrac{\rm kg}{\rm min} = \dfrac18 \dfrac{\rm kg}{\rm min}[/tex]
If [tex]A(t)[/tex] is the amount of salt (in kg) in the tank at time [tex]t[/tex] (in min), then the salt flows out of the tank at a rate of
[tex]\left(\dfrac{A(t)}{1000+(5-5)t} \dfrac{\rm kg}{\rm L}\right) \left(5 \dfrac{\rm L}{\rm min}\right) = \dfrac{A(t)}{200} \dfrac{\rm kg}{\rm min}[/tex]
The net rate of change in the amount of salt in the tank at any time is then governed by the linear differential equation
[tex]\dfrac{dA}{dt} = \dfrac18 - \dfrac{A(t)}{200}[/tex]
[tex]\dfrac{dA}{dt} + \dfrac{A(t)}{200} = \dfrac18[/tex]
I'll solve this with the integrating factor method. The I.F. is
[tex]\mu = \exp\left(\displaystyle \int \frac{dt}{200}\right) = e^{t/200}[/tex]
Distributing [tex]\mu[/tex] on both sides of the ODE gives
[tex]e^{t/200} \dfrac{dA}{dt} + \dfrac1{200} e^{t/200} A(t) = \dfrac18 e^{t/200}[/tex]
[tex]\dfrac d{dt} \left(e^{t/200} A(t)\right) = \dfrac18 e^{t/200}[/tex]
Integrate both sides.
[tex]\displaystyle \int \frac d{dt} \left(e^{t/200} A(t)\right) \, dt = \frac18 \int e^{t/200} \, dt[/tex]
[tex]e^{t/200} A(t) = \dfrac{200}8 e^{t/200} + C[/tex]
[tex]A(t) = 25 + Ce^{-t/200}[/tex]
Given that [tex]A(0)=50\,\rm kg[/tex], we find
[tex]50 = 25 + Ce^0 \implies C = 25[/tex]
so that
[tex]A(t) = 25 + 25e^{-t/200}[/tex]
Then the amount of salt in the tank after 1 hr = 60 min is
[tex]A(60) = 25 + 25e^{-60/200} = \boxed{25 \left(1 + e^{-3/10}\right)}[/tex]
