Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the power dissipated by the load resistance is given by
$$P=I^2R\tag 1$$
We need to find $I$ where we know that the inner resistance in the battery $r$ and the load resistance $R$ are in series, so
$$\varepsilon=I(r+R)$$
Hence,
$$I=\dfrac{\varepsilon}{R+r}$$
Plug into (1),
$$P=\left(\dfrac{\varepsilon}{R+r}\right)^2R $$
$$P= \dfrac{\varepsilon^2R}{(R+r)^2} \tag 1$$
Now we need to find the maximum power dissipated by the load resistance which is found as follows.
$$\dfrac{dP}{dR}= \dfrac{d}{dR}\left(\dfrac{\varepsilon^2R}{(R+r)^2}\right)=0 $$
Hence,
$$\dfrac{\varepsilon^2(R+r)^2- 2\varepsilon^2R(R+r) }{(R+r)^4}=0$$
$$\dfrac{\varepsilon^2R^2+2rR\varepsilon^2+\varepsilon^2r^2- [2\varepsilon^2R^2 +2\varepsilon^2Rr ]}{(R+r)^4}=0$$
$$\dfrac{\varepsilon^2R^2+2rR\varepsilon^2+\varepsilon^2r^2- 2\varepsilon^2R^2 -2\varepsilon^2Rr }{(R+r)^4}=0$$
$$\dfrac{ \varepsilon^2r^2- \varepsilon^2R^2 }{(R+r)^4}=0$$
Thus,
$$r^2-R^2=0$$
Hence,
$$\boxed{R=r}$$
Therefore, the power dissipated by the load resistance is maximized when it is equal to the inner resistance of the battery.
$$\color{blue}{\bf [b]}$$
As we found above, the power dissipated by the load resistance is maximized when $R=r$, plug that into (1),
$$P_{max}= \dfrac{\varepsilon^2r}{(2r)^2} =\dfrac{\varepsilon^2}{4}$$
Plug the given;
$$P_{max}= \dfrac{9^2}{4}=\color{red}{\bf 20.3}\;\rm W$$
$$\color{blue}{\bf [c]}$$
When $R\rightarrow\infty$, then $\varepsilon=I(R+r)=IR$, and hence $I\approx 0$. Thus,
$$P=I^2R=(0)^2R\approx 0$$
and when $R\rightarrow 0$, then $\varepsilon=I(R+r)=Ir$, and hence $I\approx I_{max}$. Thus,
$$P=I^2R=I_{max}^2(0)\approx 0$$
In both cases, $P$ is zero, so there must be some value of $R$ that makes $ P$ is maximum.
