Define $f(n)$ as $1$ if $n$ is odd, and $3$ if $n$ is even.
So i have $f(odd) = 1$ and $f(even) = 3$
If a function $f$ is multiplicative then if $gcd(m,n) = 1$ then $f(m * n) = f(m) * f(n)$
This seems like it is rather trival here, so im gonna give it a shot
Let $f(n) = 3$, thus $n$ is an even number
If $n$ is even, then the prime factorization of $n$,
$2^{a_{1}} * 3^{a_{2}} * 5^{a_{3}}* ... *p^{a_{k}}$ has $a_{1} > 0$ thus
$f(n) = f(2^{a_1}) * f(m)$ where $gcd(2^{a_1},m) = 1$ and $m$ is odd
Therefore $f(n) = 3 * 1 = 3$
Let $f(n) = 1$
if $f(n) = 1$ then n is odd, therefore n has all odd prime factors thus
$f(n) = f(3^{a_1}) * f(5^{a_2})... = 1 * 1 * 1 ... = 1$
I just want to know if this above is a solid proof
The second part of this question says Let $F$ be the summatory function of $f$. Calculate $F$ completely by finding $F(p^a)$ for $p$ a prime and $a$ a positive integer. Let $F(n) = \sum _{d|n} f(d)$
The only way I can think of doing this is, if p = 2, then $F(p^a) = 3a + 1$, otherwise $F(p^a) = a+1$ since if p is odd, then $F(p^a)$ will simply equal the number of divisors that divide $p^a$ and if p is even, $f(1) + 3a$
I am not sure this is right though,
The proof of multiplicativity is not entirely clear. Let $\gcd(a,b)=1$. We want to prove that $f(a,b)=f(a)f(b)$. There are $3$ cases (i) $a$, $b$ both odd; (ii) $a$ odd, $b$ even; (iii) $a$ even, $b$ odd. In Case (i), $f(a)=f(b)=1$. But $ab$ is odd, so $f(ab)=1$. It follows that $f(ab)=f(a)f(b)$. In Case (ii), $f(a)=1$ and $f(b)=3$. But $ab$ is even, so $f(ab)=3$. Thus $f(ab)=f(a)f(b)$. Case (iii) is like Case (ii).
The summatory function $F$, by general theory, is multiplicative. So if $n=p_1^{a_1}\cdots p_t^{a_t}$, where the $p_i$ are distinct primes, then we know $F(n)$ once we know the $F(p_i^{a_i})$.
We calculate $F(p^a)$ where $p$ is an odd prime. The divisors of $p^a$ are $1,p,\dots, p^a$, so there are $a+1$ of them. It follows that $F(p^a)=a+1$.
You calculated correctly $F(2^a)$.