Regarding proving a result related to $\alpha$- multiplicative function from exercises of Tom M Apostol I am trying exercises of Tom M Apostol Modular functions and Dirichlet series in number theory of Chapter-6 and I could not think about this problem. 

This problem uses concepts introduced in beginning of 1st problem but note that I have doubt only in 4 th problem. 
  I am adding its images 
Image of 4th problem 
  

My attempt - I tried putting g =1 in defination of $\alpha$ multiplicative functions and then multiplying by $\alpha$(n) but that didn't helped. 
Also I tried putting f(mnd) f(n/d) in place of f(m) f(n) in defination of $\alpha$ multiplicative function but that also doesn't yields. 

Edit 1 Also in 5 th problem I could not prove this side -->assuming f to be multiplicative and proving it to be $\alpha $ multiplicative. I tried making cases. 

Case when gcd (m, n) = 1 or m= $p^k $ and n = p are trivial. But I tried using induction on $p^k$ and $p^l$ but it doesn't solved the problem. 
Can someone please give hints for them. 
 A: $\sum_{d \mid n} \mu(d)f(mnd)f(\frac{n}{d}) = \sum_{d \mid n} \mu(d)\left(\sum_{r \mid (mnd,\frac{n}{d})} \alpha(r)f(\frac{mn^2}{r^2})\right)$. Since $\frac{n}{d} \mid mnd$, we get $\sum_{d \mid n} \sum_{r \mid \frac{n}{d}} \mu(d)\alpha(r)f(\frac{mn^2}{r^2}) = \sum_{r \mid n} \sum_{d \mid \frac{n}{r}} \mu(d)\alpha(r)f(\frac{mn^2}{r^2}) = \alpha(n)f(m)$, the last using $\sum_{d \mid k} \mu(d) = 1_{k=1}$.
Here's a hint for problem 5. Write $m = p_1^{\alpha_1}\dots p_k^{\alpha_k}q_1^{\beta_1}\dots q_t^{\beta_t}$ and $n = p_1^{\alpha_1}\dots p_k^{\alpha_k}r_1^{\gamma_1}\dots r_l^{\gamma_l}$, where $p_i,q_i,r_i$'s are primes and $\{q_1,\dots,q_t\}\cap\{r_1,\dots,r_l\} = \emptyset$. Then, by multiplicativity, $f(m)f(n) = f(p_1^{\alpha_1})^2\dots f(p_k^{\alpha_k})^2f(q_1^{\beta_1})\dots f(r_l^{\gamma_l})$. And, $$\sum_{d \mid (m,n)} \alpha(d)f(\frac{mn}{d^2}) = \sum_{\substack{\lambda_1,\dots \lambda_k \\ 0 \le \lambda_i \le \alpha_i}} \alpha(p_1^{\lambda_1})\dots \alpha(p_k^{\lambda_k}) f(p_1^{2\alpha_1-2\lambda_1})\dots f(p_k^{2\alpha_k-2\lambda_k}) f(q_1^{\beta_1})\dots f(r_l^{\gamma_l}).$$ This sum splits up into $f(q_1^{\beta_1})\dots f(r_l^{\gamma_l})\prod_i (\sum_{0 \le \lambda_i \le \alpha_i} \alpha(p_i^{\lambda_i}) f(p_i^{2\alpha_i-2\lambda_i}))$. The point is that it is sufficient to prove $$f(p^\beta)^2 = \sum_{0 \le \lambda \le \beta} \alpha(p^\lambda)f(p^{2\beta-2\lambda})$$ for any prime $p$ and $\beta \ge 1$. To prove this, you have to use the complete multiplicativity of $\alpha$ and the formula $f(p^{k+1}) = f(p)f(p^k)-\alpha(p)f(p^{k-1})$.
