Just note that $\textstyle\prod_{p\in\mathbb{P}}\left(1-\frac1p\right) = \prod_{p\in\mathbb{P}}\sum_{i\ge 0}\frac{1}{p^i} = \sum_{n\ge 1}\frac1n$.

The solution for the part $a)$ We choose n to be prime. call $n=p$, then $\frac{\sigma(\phi(n))}{\phi(\sigma(n))}$= $\frac{\sigma(p-1)}{\phi(p+1)}$. For $p>2$, $p+1$ is never a prime. So $\phi(p+1)\le p-1$. If we can prove that $\frac{\sigma(p-1)}{p-1}$ is unbounded, then we are done. For any $k$, let $lcm(1,2,..,k)=S_k$ . By Dirichlet Theorem we know that for any $k$ , there exist a prime number $q_k$ such that $q_k\equiv 1 (mod S_k)$. Now take a look at, $\sigma(q_k-1)$ , since $q_k\equiv 1 (mod S_k)$ we know that $\sigma(q_k-1)\ge (q_k-1).(1+\frac{1}{2}+...+\frac{1}{k})$. We conclude that. $\frac{\sigma(q_k-1)}{\phi(q_k+1}\ge\frac{\sigma(q_k-1)}{q_k-1}\ge 1+\frac{1}{2}+..+\frac{1}{k}$. Divergence of harmonic series gives the desired result. $\hspace{0.4cm} \blacksquare$ The solution for part $b)$ For this part of the solution we choose $n=(p_1.p_2...p_k)^{q-1}$ , where $p_1,p_2,...,p_k$ denote the first k primes starting from $2$ and $q$ is a prime number such that $q>p_k^{k}$ .($p_k$ is $k$. prime number) First, we are going to give a bound for the numerator. $Claim:$ $\sigma(\phi(n))\le n$ $Proof:$ $\phi(n)=n.\frac{p_1-1}{p_1}.\frac{p_2-1}{p_2}...\frac{p_k-1}{p_k}$ , also we know that the set of primes dividing $\phi(n)$ is a subset of $\{p_1,p_2,...,p_k\}$ , that's because the maximal prime diving an integer never increases when $\phi$ applied. Now, denote $\phi(n)=k$ for the ease of notation. $\sigma(k)=k.\prod_{\substack{p^\alpha \mid| k}} (1+\frac{1}{p}+...+\frac{1}{p^\alpha})$ , we know that for any prime $p$ and any positive integer $\alpha$ , $\frac{p}{p-1}\ge (1+\frac{1}{p}+...+\frac{1}{p^\alpha})$. So $\prod_{\substack{p^\alpha \mid| k}} (1+\frac{1}{p}+...+\frac{1}{p^\alpha})\le \prod_{\substack{p \mid k}} \frac{p}{p-1} $. Also because the $\frac{p}{p-1}>1$ and the set of primes dividing $\phi(n)$ is a subset of $\{p_1,p_2,...,p_k\}$ , $\prod_{\substack{p \mid k}} \frac{p}{p-1}\le \frac{p_1}{p_1-1}.\frac{p_2}{p_2-1}...\frac{p_k}{p_k-1}$. So, at the end we get $\sigma(\phi(n))=\sigma(k)=k.\prod_{\substack{p^\alpha \mid| k}} (1+\frac{1}{p}+...+\frac{1}{p^\alpha}) \le k. \frac{p_1}{p_1-1}.\frac{p_2}{p_2-1}...\frac{p_k}{p_k-1}$. Also $k=n.\frac{p_1-1}{p_1}.\frac{p_2-1}{p_2}...\frac{p_k-1}{p_k}$ , from this we get $ k. \frac{p_1}{p_1-1}.\frac{p_2}{p_2-1}...\frac{p_k}{p_k-1}=n$. Hence , we arrive to the desired result that $\sigma(\phi(n))\le n$ . $\hspace{0.4cm} \blacksquare$ Now , we are looking at denominator , $\sigma(n)=n.\prod_{\substack{p^\alpha \mid| n}} (1+\frac{1}{p}+...+\frac{1}{p^\alpha})$ , hence, $\sigma(n)\ge n.\frac{p_1+1}{p_1}.\frac{p_2+1}{p_2}...\frac{p_k+1}{p_k}$ . Also $\sigma(n)=\frac{p_1^{q}-1}{p_1-1}.\frac{p_2^{q}-1}{p_2-1}...\frac{p_k^{q}-1}{p_k-1}$. At this point, we are going to make an important observation about primes dividing $\sigma(n)$. For any prime $r$, if $r\mid\sigma(n) $ then $r>q$. We are going to consider two cases to prove this observation. $i)$ For a prime divisor $p$ of $\sigma(n)$ ,$r\mid p^{q}-1$ but $r\nmid p-1$. In this case, $ord_r(p)\mid gcd(r-1,q)$ , but $ord_r(p)\nmid gcd(r-1,1)$ .This suggest that $q\mid r-1$. From this we get $r\ge q+1$. $ii)$ For a prime divisor $p$ of $r\mid p-1,\frac{p^q-1}{p-1} $ If $r=2$ , then $\frac{p^q-1}{p-1} \equiv q (mod 2) $ Since $q$ is an odd prime. This is not possible. If $r$ is an odd prime , then by applying LTE , $\nu_r(\frac{p^q-1}{p-1})=\nu_r(p-1)+\nu_r(q)-\nu_r(p-1)=\nu_r(q)>0$ . So $r=q$ , but this is not possible because this suggest $q\mid p-1$ . But this is not possible because $q>p_k^k$. After this, we are going to show that number of distinct prime divisors of $\sigma(n)$ is at most $q$. That's because for any $j\in \{1,2,...,k\}$ , number of distinct prime divisors of $\frac{p_j^q-1}{p_j-1}$ is at most $\frac{q}{k}$. To show this let us call this number to be $f(j)$ then the multiplication of this different primes is at least $q^{f(j)}$ . Also $q^{q/k}>p_k^q>\frac{p_j^q-1}{p_j-1}$. That's why $f(j)\le \frac{q}{k}$. Now we call $\sigma(n)=T$ for the ease of notation. $\phi(T)\ge T.(\frac{q}{q+1})^q$. That's because, all the prime divisor of $T$ are at least $q+1$ and their total number is at most $q$. Using simple calculations we are going to show that $(\frac{q}{q+1})^q\ge 1/3$. $(q+1)^q=\sum_{k=0}^{q} q^{q-k}.\binom{q}{k}$, for every $k$, $\binom{q}{k}\le \frac{q^k}{k!}$ So $\sum_{k=0}^{q} q^{q-k}.\binom{q}{k}\le \sum_{k=0}^{q} \frac{q^{q}}{k!}$. So $(q+1)^q \le q^q(1+1+\frac{1}{2!}+...+\frac{1}{k!})\le q^q.(1+1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{2^k})\le 3.q^q$.(which gives the desired result). So $\phi(T)\ge T.(\frac{1}{3})$. Now ,we are in the last part of the proof. $\phi(\sigma(n))\ge \frac{T}{3}\ge n.\frac{p_1+1}{p_1}.\frac{p_2+1}{p_2}...\frac{p_k+1}{p_k}.\frac{1}{3} $ For any $j$ it is easy to see that $\frac{p_j+1}{p_j}\ge \frac{p_{j+1}}{p_{j+1}-1}$. So $\frac{p_1+1}{p_1}.\frac{p_2+1}{p_2}...\frac{p_k+1}{p_k}\ge \frac{p_2}{p_2-1}.\frac{p_3}{p_3-1}...\frac{p_{k+1}}{p_{k+1}-1} $. The multiplication $M_k=\frac{p_2}{p_2-1}.\frac{p_3}{p_3-1}...\frac{p_{k+1}}{p_{k+1}-1} $ goes to infinity as $k$ gets bigger. That's because $\frac{p}{p-1}=1+\frac{1}{p}+\frac{1}{p^2}+...$ , So the multiplication $M_k=\frac{p_2}{p_2-1}.\frac{p_3}{p_3-1}...\frac{p_{k+1}}{p_{k+1}-1} $ is bigger than the sum of the reciprocal all odd number from 1 to $k$. (All number's smaller than equal to k has its all prime divisor smaller than $p_{k+1}$). This is divergent because harmonic series diverge. Now if we look at $\frac{\sigma(\phi(n))}{\phi(\sigma(n))}$, $\frac{\sigma(\phi(n))}{\phi(\sigma(n))}\le \frac{n}{n.M_k.\frac{1}{3}}=\frac{3}{M_k}$ . The divergence of $M_k$ gives the desired result for the part b). $\hspace{0.4cm} \blacksquare$