Answer
a) $V_1$ increases
$V_2$ decreases
$V_3$ drops to 0
b) $I_1$ increases
$I_2$ decreases
$I_3$ drops to 0
c) $\mathcal{E}$ increases
d) $V_{ab}=8.49V$
e) $V_{ab}=8.61V$
Work Step by Step
Lets name resistors 1 to 3 from left to right
a) Switch closed
$R_{eq}=R+r+\frac{1}{\frac{2}{R}}=1.5R+r$
$I_{closed}=\frac{V}{R_{eq}}=\frac{\mathcal{E}}{1.5R+r}$
Switch open
$R_{eq}=R+R+r=2R+r$
$I_{open}=\frac{V}{R_{eq}}=\frac{\mathcal{E}}{2R+r}$
When the switch is opened, the current decreases. This means the voltage across $R_1$ decreases. The previously divided current through $R_2$ and $R_3$ now go through $R_2$. This means the voltage across $R_2$ increases. Lastly, no current flows through $R_3$ so the voltage across it drops to 0.
b) $I=\frac{V}{R}$, so current is proportional to V. If V increases, I increases and vice versa
c) $V_{ab}=\mathcal{E}-Ir$
When the current is decreased, the right hand term $Ir$ decreases and the terminal velocity increases.
d) $I_{closed}=\frac{9.0V}{(1.5)(5.50\Omega)+0.50\Omega}=1.03A$
$V_{ab}=\mathcal{E}-Ir=9.0V-(1.03A)(0.50\Omega)=8.49V$
e) $I_{open}=\frac{9.0V}{2(5.50\Omega)+0.50\Omega}=0.783A$
$V_{ab}=\mathcal{E}-Ir=9.0V-(0.783A)(0.50\Omega)=8.61V$