(a) We will prove (a) by induction when $n=1$ we set $a=a_1$ and so it works when $n=2$ $(1+\frac{1}{a_1})(1+\frac{1}{a_2})=1+\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_1a_2}=1+\frac{1}{a}$ and so $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_1a_2}=\frac{1}{a}$ $a_1+a_2+1=\frac{a_1a_2}{a}$ $aa_1+aa_2+a=a_1a_2$ $a_1a_2-aa_1-aa_2-a=0$ $(a_1-a)(a_2-a)=a^2+a$ we can let $a_1=2a$ and $a_2=2a+1$ and we are done Let us assume that $n=k$ works we know we can show $1+\frac{1}{a} = (1+\frac{1}{a})(1+\frac{1}{b})$ as we proved the case for n=2 we also know that $(1+\frac{1}{a})$ can be shown as $\prod_{n=1}^k(1+\frac{1}{a_i})$ (by induction hypothesis) and so $1+\frac{1}{a}=\prod_{n=1}^k(1+\frac{1}{a_i})\times(1+\frac{1}{b})$ and so we proved it works for $n=k+1$ too

(b) sketch of solution: let's assume that $a_1<a_2< \cdots <a_n$ $$ (1+ \frac{1}{a})= \Pi_{i=1}^n (1+ \frac{1}{a}) $$here on of the $ (1 + \frac{1}{a_i})$ must be greater than $\sqrt[n]{ 1+ \frac{1}{a}} $ hence the values of $a_1$ is bounded. now consider $\frac { 1+ \frac{ 1}{a}}{1+ \frac{1}{a_1}}$ and using the previous trick to bound the value of $a_2$ for every value of $a_1$ and so on. in this solution i just prove that this is finite.