Proving exponential distribution given a function 
I'll admit that I have no idea how to work this one out. I have gotten this far:
I know that I don't have to use a differential equation. This is a hint intended to help:
$$\frac{f(a+h)-f(a)}{h}=\frac{f(a)f(h)-f(a)}{h}
=f(a)\frac{f(h)-1}{h}=f(a)\frac{f(h)-f(0)}{h}$$$$\text{in the limit }h\to \infty, $$$$=f(a)f'(0)$$$$f'(a)=f(a) c\to\frac{f'(a)}{f(a)}=c$$
How this hint helps is a mystery to me; any help would be wonderful.
 A: The hint show that:
$$f'(a)=f(a) f'(0)$$
for any $a \in (0,+\infty)$. In other words with $-\lambda=f'(0)$ you have:
$$\forall x \in (0,+\infty), \, f'(x)=-\lambda f(x) \tag{1}$$
More over $f(0)=f(0+0)=f(0)^2$ so:
$$f(0)=1 \tag{2}$$
There is only one differentiable function satisfying $(1)$ and $(2)$ and it is $x \mapsto e^{-\lambda x}$.
A: The challenge in the question is to avoid using differential equations (or related theory) in the solution.  It turns out an insightful demonstration can be achieved by assuming much less than is originally supposed and that it requires minimal knowledge of Calculus.

Let us assume only that $f$ is bounded and bounded away from zero on the interval $(0,1].$  (This means there exist $u \ge l \gt 0$ and for all $x\in (0,1],$ $l\le f(x) \le u.$) 
Since $f(1)\ne 0$ and the exponential function $\lambda\to e^\lambda$ attains all positive real values, there exists a number $\lambda$ for which 
$$e^\lambda f(1) = 1.$$
Define the function $\phi:[0,\infty)\to(0,\infty)$ via
$$\phi(x) = e^{\lambda x} f(x).$$
Because the function $x\to e^{\lambda x}$ is bounded and bounded away from zero on $(0,1],$ $\phi$ itself is bounded and bounded away from zero on $(0,1].$  In particular,
$$\phi(1) = e^{\lambda 1} f(1) = 1.$$
I will show that $\phi$ is constantly $1$.
The defining relation for $f$ implies that for all $a,b\gt 0,$
$$\phi(a+b) = e^{\lambda(a+b)}f(a+b) = e^{\lambda a} f(a)\, e^{\lambda b} f(b) = \phi(a)\phi(b).$$
In case $b=1,$ this yields
$$\phi(a+1) = \phi(a)\phi(1) = \phi(a)1 = \phi(a).$$
By an easy induction, this result (which shows $\phi$ is periodic with period $1$ and therefore the values it attains are precisely the values it attains between $0$ and $1$) demonstrates $\phi$ is everywhere bounded and bounded away from zero.
Suppose $a \gt 0$ is any number for which $\phi(a) \ne 1.$  By induction on the natural number $n,$
$$\phi(x + na) = \phi(x)\phi(a)^n.$$
As $n$ grows, the right hand side either diverges (when $\phi(a) \gt 1$) or converges to $0$ (when $\phi(a) \lt 1$) because $\phi(x) \ne 0.$  That contradicts the previous conclusion that $\phi$ is bounded and bounded away frmo zero.  It follows that such a number $a$ cannot exist; that is, the only value $\phi$ attains is $1$.
Recalling the definition of $\phi,$ this means $1 = e^{\lambda x}f(x).$  Solving for $f$ gives
$$f(x) = e^{-\lambda x}$$
for all $x \gt 0.$

Part (b) of the question is incorrect.  However, if in addition you assume $\Pr(X\gt 0) \gt 0,$ then the foregoing analysis applies to $f(x) = \Pr(X \gt x)$ because the axioms of probability imply $f$ is bounded above by $1$ and below by $\Pr(X\gt 0).$ We conclude there is a real number $\lambda$ for which whenever $x\gt 0,$
$$\Pr(X \gt x) = e^{-\lambda x}.$$
Necessarily $\lambda \gt 0$ (for otherwise this function would not decay towards $0$ as $x$ grows).  This is the defining formula for an Exponential$(\lambda)$ distribution, QED.
