Calculating pH of rain water
The value $\pu{1.3653 mol L^{−1} atm^{−1}}$ is the solubility constant (or Henry's law solubility constant), not the solubility. The solubility is defined as the maximum possible concentration (the saturation concentration) of a solute under given solution conditions (e.g. temperature and pressure), whereas the solubility constant $H^{cp}$ defines how solute partitions between the gas (concentration in gas here given as the partial pressure in atmospheres — although this is not stated in the OP it can be inferred from the expected answer) and solution phases (concentration in $\pu{mol/L}):$

$$H^{cp} = \frac{c}{p}= \pu{1.3653 mol L^{−1} atm^{−1}}$$  

Assuming atmospheric pressure of 1 atm then $\pu{10 ppm}$ translates into a partial pressure of $\pu{1e-5 atm}$ (if the atmospheric pressure is $p_T$, then the partial pressure of $\ce{SO2}$ is $p = p_T \cdot 10^{-5}$) and a solution concentration $\pu{1.365e-5 M}$  $\;(c=H^{cp} p)$. Proceeding from the definition of $\mathrm{p}K_\mathrm{a}$ the degree of dissociation $(\alpha=0.99886)$ can be computed and from this the $\mathrm{pH} = 4.865,$ which agrees with the expected result.  

You should check solubility limits when evaluating this sort of problem, but since the solubility is not given you can ignore this and assume that the limit has not been reached.