Given,    
 $$\begin{gathered} C\mathrm{harge}, q_1 = 2 \times {10}^{-7}\mathrm{C}, \\ C\mathrm{harge}, {\mathrm{q}}_2 = 3 \times {10}^{-7}\mathrm{C} \\ \mathrm{r} = 30 \mathrm{cm} = 0.3 \mathrm{m} \end{gathered}$$

where, r is the distance between two charges.

Using the formula,

$\therefore$                $\boxed{\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}}$

                          $$\begin{gathered} = \frac{9\times {10}^9\times 2\times {10}^{-7}\times 3\times {10}^{-7}}{(0.3{)}^2} \\ = 6 \times {10}^{-3} \mathrm{N} \left(\mathrm{Repulsive}\right) \end{gathered}$$

Repulsive in nature since both charges are positive.

Given,

 $$\begin{gathered} \mathrm{Dipole} \mathrm{moment}, \mathrm{p} = 4 \times {10}^{-9} \mathrm{Cm}; \\ \mathrm{\theta } = 30° \\ \mathrm{Eelectric} \mathrm{field}, \mathrm{E} = 5 \times {10}^4 {\mathrm{NC}}^{-1} \end{gathered}$$

Torque is given by  

      $$\begin{gathered} \tau =\boxed{\mathrm{p}\times \overrightarrow{\mathrm{E}}} = \mathrm{p}. \mathrm{E} \sin \mathrm{\theta } \\ = 4 \times {10}^{-9} \times 5 \times {10}^4 \times \sin 30° \end{gathered}$$
    
                $\tau =4\times {10}^{-9}\times 5\times {10}^4\times \frac{1}{2}$
                   $\tau ={10}^{-4} \mathrm{Nm}$  

Hence, two field lines never cross each other.

(a) Given,

   $\mathrm{Force} -\mathrm{F} = 0.2 \mathrm{N}$   
                 $$\begin{gathered} \mathrm{charge} -{\mathrm{q}}_1 = 0.4 \mathrm{\mu C} = 0.4 \times {10}^{-6}\mathrm{C} \\ \mathrm{charge}-{\mathrm{q}}_2 = 0.8 \mathrm{\mu C} = 0.8 \times {10}^{-6}\mathrm{C} \end{gathered}$$   

Now using the formula,                                                                   $\mathrm{F} = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Hence,                     $\boxed{{\mathrm{r}}^2 = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{\mathrm{F}}}$
                       $$\begin{gathered} {\mathrm{r}}^2 = \frac{9\times {10}^9\times 0.4\times {10}^{-6}\times 0.8\times {10}^{-6}}{0.2} \\ {\mathrm{r}}^2 = 36 \times 4 \times {10}^{-4} = 144 \times {10}^{-4} \\ \mathrm{r} = 12 \times {10}^{-2}\mathrm{m} = 12 \mathrm{cm}. \end{gathered}$$

where, r is the distance between two spheres.

(b) Force on the second sphere due to the first is same, i.e., 0.2 N because the charges in action are same and force is attractive as charges are unlike in nature.