Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the power supplied by the battery as the capacitor is being charged is given by
$$P_{\rm battery}=I\varepsilon\tag 1$$
And we know for a charging capacitor that
$$Q=Q_{\rm max}\left(1-e^{-t/RC}\right)$$
Recalling that $I=dQ/dt$, so
$$I=\dfrac{d\left[ Q_{\rm max}\left(1-e^{-t/RC}\right)\right]}{dt}= \dfrac{Q_{\rm max}}{RC}e^{-t/RC}$$
where $Q_{\rm max}=\varepsilon C$,
$$I= \dfrac{\varepsilon C }{RC}e^{-t/RC}$$
$$I= \dfrac{\varepsilon }{R }e^{-t/RC}\tag 2$$
Plug $I$ from (2) into (1);
$$P_{\rm battery}=\dfrac{\varepsilon^2 }{R }e^{-t/RC} $$
Now we need to find the total energy supplied by the battery, where
$P=dE/dt$
Hence,
$$\dfrac{dE }{dt}=\dfrac{\varepsilon^2 }{R }e^{-t/RC} $$
$$ dE =\dfrac{\varepsilon^2 }{R }e^{-t/RC} dt $$
Integrating both sides; where we need $t$ after a very long time so $t$ is from 0 to $\infty$
$$\int_0^E dE =\int_0^\infty\dfrac{\varepsilon^2 }{R }e^{-t/RC} dt $$
$$ E = \dfrac{\varepsilon^2 }{R }(-RC)\bigg[e^{-t/RC} \bigg]_0^\infty=\dfrac{\varepsilon^2 }{R }(-RC)(0-1)$$
$$ \boxed{E_{\rm battery} = \varepsilon^2C } $$
$$\color{blue}{\bf [b]}$$
We know that the power dissipated by the resistor is given by
$$P_{R}=I^2R$$
Plug from (2),
$$P_{R}=\left[ \dfrac{\varepsilon }{R }e^{-t/RC}\right]^2R= \dfrac{\varepsilon^2 }{R }e^{-2t/RC}$$
where, again, $P=dE/dt$, so
$$\dfrac{dE}{dt}=\dfrac{\varepsilon^2 }{R }e^{-2t/RC}$$
Hence,
$$dE=\dfrac{\varepsilon^2 }{R }e^{-2t/RC}dt$$
Integrating both sides; where we need $t$ after a very long time so $t$ is from 0 to $\infty$
$$\int_0^EdE=\dfrac{\varepsilon^2 }{R }\int_0^\infty e^{-2t/RC}dt$$
$$ E=\dfrac{\varepsilon^2 }{R }\dfrac{-RC}{2} \bigg[e^{-2t/RC}\bigg]_0^\infty$$
$$ \boxed{E_{R}=\frac{ 1 }{2 } \varepsilon^2 C }$$
$$\color{blue}{\bf [c]}$$
The energy stored in the capacitor is given by
$$E=\dfrac{Q^2_{\rm max}}{2C}$$
where $Q_{\rm max}=\varepsilon C$
$$E=\dfrac{\varepsilon^2 C^2}{2C}$$
$$\boxed{E_C=\dfrac{1}{2 }\varepsilon^2 C }$$
$$\color{blue}{\bf [d]}$$
To see if the energy is conserved or not, we need to make sure that the sum of the energy stored in the capacitor plus the energy dissipated in the resistor is equal to the energy supplied by the battery.
And from above, yes the energy is conserved.
