Since $f(2)=5$ and $f(mn)=f(m)+f(n)$, then $f(4)=f(2 \cdot 2)=f(2)+f(2)=10$. Since $f(3)=7$, then $f(12)=f(4 \cdot 3)=f(4)+f(3)=10+7=17$. While this answers the question, is there actually a function that satisfies the requirements? The answer is yes. One function that satisfies the requirements of the problem is the function $f$ defined by $f(1)=0$ and $f\left(2^{p} 3^{q} r\right)=5 p+7 q$ for all non-negative integers $p$ and $q$ and all positive integers $r$ that are not divisible by 2 or by 3. Can you see why this function satisfies the requirements?