Exponential or a constant function 
Here f has to be exponential function or constant function f(x)=1. Option one and two will contradict a^x and f(x)=1. Option 3 is true. Here f is never zero.
Confused whether option 4 is true
 A: If the function satisfies (3), it has to satisfy (4). Take $r＝\frac{1}{2}$, then $f(x)$ has to be positive.
A: $f(2x)=f(x)^2$. Thus $f(y)\geq0$ for all $y$. If $f(x_0)=0$ for some $x_0$ then $f(x)=f((x-x_0)+x_0)=f(x-x_0) f(x_0)=0$ for all $x$ contradicting that $f(x)$ tends to 1 for $x\to0$. Thus $f(x)>0$ for all $x$ and therefore by $f(x)=f(x+0)=f(x)f(0)$ also $f(0)=1$. This shows that $f$ is continuous at 0 and then everywhere since $f(x+y)=f(x)f(y)$. Taking logarithm shows that
$g:=\ln\circ f:\mathbb{R}\to\mathbb{R}$ is a continuous homomorphism. This implies that there is some constant $\alpha$ such that $g(x)=\alpha x$ for all $x$. Accordingly $f(x)=\exp(\alpha x)=a^x$ for $a:=\exp(\alpha)$. Thus also 3) is true, even for all real numbers $r$.
A: 3) and 4) are true. 1) and 2) are false. In fact we can show that $f(x)=e^{cx}$ for some real number $c$. If $f(x)=0$ for some $x$ then $f(y)=f(x)f(y-x)=0$ for all $y$ but this contradicts the hypothesis. Since $f(x)=f(x/2)^{2} \geq 0$ it follows that $f>0$. Let $g(x)=\ln f(x)$ Then $g(x+y)=g(x)+g(y)$ and $g(x) \to 0$ as $x \to 0$. By a well known result this implies $g(x)=cx$ for som ereal number $c$. Hence $f(x)=e^{cx}$. Conversely a function of this type satisfies teh functional equation. Taking $c<0$ we see that 1) and 2) false. 3) is now obvious.
