Step 1: Part (i)(I): why the current starts at once:
The signal that decides the motion of the electrons is the electric field, not the electrons themselves. The field spreads through the wire at a speed close to that of light.
Each free electron in the wire gets pushed as soon as the field reaches it. All of them begin to drift together, so no waiting time is seen.
A slow drift speed only means that one electron takes long to cross the wire. It says nothing about how fast the current is set up.
Step 2: Part (i)(II): small r for a large current:
Terminal voltage is $V = \varepsilon - Ir$. The lost voltage $Ir$ grows with the current drawn.
For a low-voltage supply, say a few volts, a high current such as 100 A would eat up the whole emf if $r$ were even 0.1 ohm or more.
So $r$ must be very small, to keep $V$ close to $\varepsilon$ while a high current flows.
Step 3: Part (i)(III): meaning of Ohm's law:
Ohm's law is a claim about a material: the ratio $V/I$ stays constant when $V$ is changed (temperature fixed). It gives a straight-line V-I graph through the origin.
The equation $V = IR$ is only the definition of resistance. It is true for any conductor at any instant, even when $V/I$ changes with $V$ or depends on the direction of the current, as in a p-n junction diode.
So writing $V = IR$ does not by itself assert Ohm's law.
Step 4: Part (ii): use conductances:
Cells in parallel can be combined using the reciprocal of the internal resistance. The data are $\varepsilon_1 = 12$ V, $r_1 = 1\ \Omega$, $\varepsilon_2 = 6$ V, $r_2 = 0.5\ \Omega$, with like poles joined together.
The formulas are
\[ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}, \qquad \frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \]
Step 5: Put in the numbers:
\[ \frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{0.5} = 1 + 2 = 3\ \Omega^{-1} \Rightarrow r_{eq} = \frac{1}{3}\ \Omega \approx 0.33\ \Omega \]
\[ \frac{\varepsilon_{eq}}{r_{eq}} = \frac{12}{1} + \frac{6}{0.5} = 12 + 12 = 24\ \text{A} \]
\[ \varepsilon_{eq} = 24 \times \frac{1}{3} = 8\ \text{V} \]
Step 6: Quick check with the short-circuit current:
If A and B are joined by a wire, the 12 V cell gives 12 A and the 6 V cell gives 12 A. The total is 24 A.
The equivalent cell gives $8/(1/3) = 24$ A. The two results are the same.
Final Answer:
The field reaches every electron almost at once, a high current needs a very small internal resistance, and V = IR is a definition of resistance. The equivalent cell has emf 8 V and internal resistance 1/3 ohm.
\[ \boxed{\varepsilon_{eq} = 8\ \text{V},\quad r_{eq} = \frac{1}{3}\ \Omega \approx 0.33\ \Omega} \]