2. For this, you need to know the Cauchy product theorem, which says that if $\sum_{m=0}^\infty a_m$ and $\sum_{n=0}^\infty b_n$ are absolutely convergent series then their product is given by $\Bigl(\sum_{m=0}^\infty a_m\Bigr)\Bigl(\sum_{n=0}^\infty b_n\Bigr) = \sum_{n=0}^\infty \Bigl(\sum_{k=0}^n a_kb_{n-k}\Bigr)$.
If you apply that to the power series for exp(a) and exp(b) then you get $\Bigl(\sum_{m=0}^\infty \frac{a^m}{m!}\Bigr)\Bigl(\sum_{n=0}^\infty \frac{b^n}{n!}\Bigr) = \sum_{n=0}^\infty \Bigl(\sum_{k=0}^n \frac{a^kb^{n-k}}{k!(n-k)!}\Bigr)$. We want to show that this is equal to $\exp(a+b) = \sum_{n=0}^\infty \frac{(a+b)^n}{n!}$. Comparing the two series, you see that this means showing that $\frac{(a+b)^n}{n!} = \sum_{k=0}^n \frac{a^kb^{n-k}}{k!(n-k)!}$. But if you multiply both sides by n!, that is just the binomial expansion of $(a+b)^n$.