Let's denote the event of getting a sum of 5 be represented as 5 and that of 7 by 7
hence 5 can be obtained by the combination {1,4 and 2,3} and 7 by {1,6  2,5  3,4} 
thus $\mathrm{P}\left(5\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)[\mathrm{P}\left(1\right)\mathrm{P}\left(4\right)+\mathrm{P}\left(2\right)\mathrm{P}\left(3\right)]=2\times [\frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}]=\frac{1}{9}$

Similarly $\mathrm{P}\left(7\right)=\left(\frac{2}{1}\right)[\mathrm{P}\left(1\right)\mathrm{P}\left(6\right)+\mathrm{P}\left(2\right)\mathrm{P}\left(5\right)+\mathrm{P}\left(3\right)\mathrm{P}\left(4\right)]$

$=2\times [\frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}]=\frac{1}{6}$

$\mathrm{P}(\overline{5}\cap \overline{7})=1-\mathrm{P}(5\cup 7)=1-\mathrm{P}\left(5\right)-\mathrm{P}\left(7\right)+0$

$=1-\frac{1}{9}-\frac{1}{6}=\frac{13}{18}$
now Let the desired event be represented by E
then $\mathrm{P}\left(\mathrm{E}\right)=\mathrm{P}\left(5\right)+\mathrm{P}(\overline{5}\cap \overline{7})\mathrm{P}\left(5\right)+\mathrm{P}(\overline{5}\cap \overline{7})\mathrm{P}(\overline{5}\cap \overline{7})\mathrm{P}\left(5\right)+........................\mathrm{\infty }$
clearly this is the infinite sum of a G.P with first term = P(5) and common ratio = $\mathrm{P}(\overline{5}\cap \overline{7})$
⇒ $\mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{P}\left(5\right)}{1-\mathrm{P}(\overline{5}\cap \overline{7})}=\frac{{\displaystyle \frac{1}{9}}}{1-{\displaystyle \frac{13}{18}}}=\frac{2}{5}$