Say that c_1,c_2!=0

We want to solve,

c_1e^{r_1 t}+c_2e^{r_2 t}=0

Divide by exp(r_2 t) to get,

c_1e^{(r_1-r_2)t}+c_2 = 0

Thus,

c_1e^{(r_1-r_2)t}=-c_2

Hence,

e^{(r_1-r_2)t}=-c_2/c_1

An exponentaial always have at most one root.

Same idea.(b) Show that x'(t) is 0 at most once

Differenciating still keeps it an exponential.