$a+b|5a+3b\implies a+b|2a$. However, $a+b>\frac{2a}{2}$, so we must have $a+b=2a\implies a=b$.

If $5a+3b$ is divisible by $a+b$, then $2a$ must be divisible by $a+b$. The only possible way for $a+b$ to be a factor of $2a$ is to be $2a$ itself. So $a=b$.

Suppose $a=dx,b=dy$, $\gcd(a,b)=d$ and $\gcd(x,y)=1$. We have $a+b | (5a+3b)-(3a+3b)=2a$ then $d(x+y)|2dx$. Since $\gcd(x,x+y)=1$, so $x+y|2$ which gives us $(x,y)=(1,1)$, then $a=b$.