Taking $p = q = 3$ in the second equation gives $f(6) = 2f(3)$, while taking $a = 2, b = 3$ in the first gives $f(6) = f(2)f(3)$. Since $f(3) \ne 0$, we have $f(2) = 2$. Next, taking $a = 3, b = 4$ in the first equation gives $f(12) = f(3)f(4)$. Taking $p = q = 2$ in the second equation gives $f(4) = 2f(2)$, so $f(12) = 2f(2)f(3) = 4f(3)$. However, we also have \[f(12) = f(5 + 7) = \] \[[f(2) + f(3)] + [f(2) + f(5)] = \] \[[2 + f(3)] + [2 + 2 + f(3)] = 6 + 2f(3)\] Then since $4f(3) = 6 + 2f(3)$, $f(3) = 3$. Then $f(5) = 5$ and $f(7) = 7$. Now, notice $f(14) = f(11) + f(3) = f(11) + 3$. But $f(14) = f(2)f(7) = 14$, so $f(11) = 11$. Applying the second rule again, $f(13) = 13$. Then \[f(2002) = f(2)f(7)f(11)f(13) = 2 \cdot 7 \cdot 11 \cdot 13 = 2002\] So $f(1999) + f(3) = f(1999) + 3 = 2002$, meaning $f(1999) = 1999$ as desired.

$f(2p)=f(p)+f(p)\to f(2p)=2f(p)$. Take $p$-odd prime, then $f(2)=2\to f(4)=4$. $f(1)=1$ from first condition. Let $f(3)=a_3,f(5)=a_5,f(7)=a_7$. $f(30)=f(2)*f(3)*f(5)=2a_3*a_5=f(13)+f(17)=f(20)-f(7)+f(20)-f(3)=8*a_5-a_3-a_7$ or 1) $2a_3*a_5=8a_5-a_3-a_7,$ 2)$f(10)=2a_5=a_3+a_7$, From (1),(2) we get $2a_3a_5=6a_5\to 2a_3=6\to a_3=3$. 3)$f(12)=4a_3=a_5+a_7$. From (2),(3) and $a_3=3$ we get $a_5=5,a_7=7$. $f(8)=f(3)+f(5)=8$. Therefore for $n\le 2^3$ we get $f(n)=n$. Now we prove, that for $n\le 2^{k+1}$ $f(n)=n$. For even $2^k<m<2^{k+1}$ it is obviosly because $m=m_1*2^l$ and $1<m_1<2^k$ - odd. Let any even number can be represented as sum of 2 distinkt primes (it cheked for all $m<10^{18}$). We use only $m=2^{k+1}-2=p_1+p_2,p_1>p_2$. Then $p_1>2^k$. Therefore $f(p_1)=f(2^{k+1}-2)-f(p_2)=p_1$. If $p=2^{k+1}-3$ or $p=2^{k+1}-1$, then $f(p)=f(2*(2^k+1))-5$ or $f(p)=f(2*(2^k+1))-3$. If $2^k+1$ is prime we prove $f(2^k+1)=2^k+1$. If $2^k+1$ is not a power of prime, then $f(2^k+1)=f(u)*f(v), (u,v)=1$, therefore $f(2^k+1)=2^k+1$. Unique solution $2^k+1=q^l=9$. For these case $f(14)=2*a_7=14=f(11)+3\to f(11)=11\to f(18)=2f(9)=f(11)+f(7)=18\to f(9)=9$. Therefore for $n<2^{k+1}$ we proved $f(n)=n$ if $n\not =p^l>2^k$. For these case $f(2*p^l)=f(q_1)+f(q_2)=q_1+q_2$. State case $n=2^{k+1}=q_1+q_2=2^{k+1}$. Therefore $f(n)=n$ for all $n<N$. We use only weakly Goldbax conjecture, that $2^{k+1}=q_1+q_2, 2^{k+1}-2=q_1+q_2, 2*p^l=q_1+q_2, l>1$. It cheked for all $n<10^{18}.$

I used the same substitutions for $f(2)$ and $f(3)$ $f(2010)=f(11)+f(1999)$. We compute $f(2010)=f(5*2*67*3)=f(5)f(2)f(3)f(67), f(67)+f(3)=f(70)=f(7)f(2)f(5)=7*5*2=70$ (It is easy to prove that $f(7)$ and $f(5)$ are $7$ and $5$ respectively) $\Longrightarrow f(67)=67 \Longrightarrow f(2010)=2010$.Now $f(9)=f(7)+f(2)=9 \Longrightarrow f(18)=18, f(7)+f(11)=f(18) \Longrightarrow f(11)=11 \Longrightarrow f(1999)=1999$

$f(2p)=f(2)f(p)=2f(p)\implies f(2)=2$ now $f(12)=f(4)f(3)=f(5)+f(7)=2f(3)+6$ so $f(3)=3,f(4)=4$ now $f(1999)=f(3)f(667)-2$ also $f(667)=f(3)[f(25)f(9)-2]-2$ now $f(9)=f(7)+2=9$ and $f(25)=\frac {f(3)f(11)+f(17)}{2}$ from $f(21)=4+f(17)=f(3)f(7)$ we get $f(21)=21$ and also $f(22)=2f(11)=f(13)+f(9)=f(11)+11$ so $f(11)=11$ and hence done.