Maximum current is drawn from a battery when the external resistance in the circuit is zero i.e., R = 0.

Given,
E.m.f of the battery, $\mathrm{\epsilon } = 12 \mathrm{V}$
Internal resistamce of the battery, $\mathrm{r} = 0.4$ 
Therefore, using the formula for maximum current drawn we get,
                           $$\begin{gathered} \boxed{{\mathrm{I}}_{\max } = \frac{\mathrm{\epsilon }}{\mathrm{r}}} \\ = \frac{12}{0.4} \\ = 30 \mathrm{A} \end{gathered}$$

Given,
EMF of battery, $\mathrm{\epsilon } = 10 \mathrm{V}$ 
Internal resistance of battery, r= $3 \mathrm{\Omega }$ 
Current flowing in the circuit, I= 0.5 A 

Using the forlmula   $\boxed{\mathrm{I} = \frac{\mathrm{\epsilon }}{\mathrm{R}+\mathrm{r}}}$ 

or                           $\mathrm{R} = \frac{\mathrm{\epsilon }}{\mathrm{I}}-\mathrm{r}$ 
where, R is the external resistance.
$\mathbf{\therefore }$       $\mathrm{R} = \frac{10}{0.5}-3 = 17 \mathrm{\Omega }$ is the required resistance.

Terminal voltage,  $$\begin{gathered} \mathrm{V} = \mathrm{IR} \\ = 0.5 \times 17 \\ = 8.5 \mathrm{V} \end{gathered}$$
                                

Given, three resistors of resistances 1$\mathrm{\Omega }, 2\mathrm{\Omega } \mathrm{and} 3\mathrm{\Omega }$ combined in series.

Given, three resistors 2$\mathrm{\Omega }, 4\mathrm{\Omega } \mathrm{and} 5\mathrm{\Omega }$ are combined in parallel. 
 
Therefore,
Total resistance of parallel combination,
                      $$\begin{gathered} \boxed{\frac{1}{\mathrm{R}}= \frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}} \\ \mathrm{or} \frac{1}{\mathrm{R}} = \frac{1}{2}+\frac{1}{4}+\frac{1}{5} \\ = \frac{10+5+4}{20} \\ = \frac{19}{20}\mathrm{\Omega } \end{gathered}$$
                           
                             $\mathrm{R} = \frac{20}{19}\mathrm{\Omega }$ 

ii) The combination is connected to a battery of 20 V.
Let the current through resistances 2Ω, 4Ω and 5Ω are I₁, I₂ and I₃ respectively.
Now, using Ohm's law across each resistor we get,
                    $$\begin{gathered} {\mathrm{I}}_1 = \frac{\mathrm{V}}{{\mathrm{R}}_1} = \frac{20}{2} = 10 \mathrm{A} \\ {\mathrm{I}}_2 = \frac{\mathrm{V}}{{\mathrm{R}}_2} = \frac{20}{4} = 5 \mathrm{A} \\ {\mathrm{I}}_3 = \frac{\mathrm{V}}{{\mathrm{R}}_3} = \frac{20}{5} = 4 \mathrm{A} \end{gathered}$$  
Hence,
Total current is given by
             $$\begin{gathered} \mathrm{I} = {\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3 \\ = 10+5+4 \\ = 19\mathrm{A}. \end{gathered}$$