Answer
${\bf 0.50}\;\Omega$, ${\bf 9.0}\;\rm V$
Work Step by Step
Let's assume that $R_1=5\;\Omega$ and $R_2=10\;\Omega$.
When the switch is open, the current is given by Ohm's law,
$$\varepsilon=I_AR_{eq}$$
where $R_{eq}=r+R_1$ where we assume that the Ammeter has negligible resistance. Hence,
$$\varepsilon-I_A(r+R_1)=0$$
where $I_A$ is the current that goes through $R_1$ and it is the same current that comes out from the battery and goes through $r$.
$$\varepsilon-I_Ar-I_AR_1=0\tag 1$$
When the switch is closed: by using Kirchoff's loop law, starting from the lower left corner and moving clockwise along the left small loop.
$$\varepsilon-Ir-I_A'R_1=0\tag 2$$
where $I_A$ is the current that goes through $R_1$, $I_2$ is the current that goes through $R_2$, and $I$ is the total current that comes out from the battery and goes through $r$.
Hence,
$$I=I_A'+I_2\tag 3$$
We can find $I$ by knowing that the two resistors $R_1$ and $R_2$ are in parallel, so they have the same voltage.
Thus,
$$I'_AR_1=I_2R_2$$
Hence,
$$I_2=\dfrac{I'_AR_1}{R_2}=\dfrac{(1.565)(5)}{(10)}=\bf 0.7825\;\rm A$$
Plug into (3)
$$I=I_A'+I_2=1.565+0.7825=\bf 2.3475\;\rm A$$
Subtract (1) from (2),
$$\varepsilon-I r-I'_AR_1 -(\varepsilon-I_Ar-I_AR_1)=0-0=0 $$
$$\varepsilon-I r-I'_AR_1 -\varepsilon+I_Ar+I_AR_1 =0 $$
$$ -I r-I'_AR_1 +I_Ar+I_AR_1 =0 $$
Now we have one unknown which is $r$ ;
$$(I_A -I)r=I'_AR_1 -I_AR_1 $$
$$r=\dfrac{I'_AR_1 -I_AR_1 }{I_A -I} $$
Plug the known;
$$r=\dfrac{(1.565)(5) -(1.636)(5)}{1.636-2.3475} $$
$$r=\color{red}{\bf 0.50}\;\Omega$$
Plug into (1) and solve for $\varepsilon$;
$$\varepsilon=I_Ar+I_AR_1 =(1.636)(0.499)+(1.636)(5)$$
$$\varepsilon=\color{red}{\bf 9.0}\;\rm V$$
