1. We will prove by induction that every positive integer $k$ can be written as sum of several distinct Fibonacci numbers.
The case $k=1$ is obvious, we have $1=f_{1}$.
Assume that every positive integer less than $n$ can be written as sum of several distinct Fibonacci numbers, consider $k=n$.
We can find $f_{i}$ is the greatest Fibonacci number which is less than $n$, we have
$$n=f_{i}+\left(n-f_{i}\right) .$$
According to the induction hypothesis, $n-f_{i}$ is a positive integer less than $n$, therefore it can be written as
$$n-f_{i}=f_{j_{1}}+f_{j_{2}}+\cdots+f_{j_{t}},\ j_{1}<j_{2}<\ldots<j_{t} .$$
We must have $i>j_{t}$ because otherwise, $n=f_{i}+f_{j_{t}}+\cdots>f_{i}+f_{i-1}=f_{i+1}$, contradiction to the choice of $i$.
So, $n=f_{j_{1}}+f_{j_{2}}+\cdots+f_{j_{t}}+f_{i}$, which is a representation of $n$ as sum of distinct Fibonacci numbers, we are done.

2. According to the part 1., an integer $n$ can be written as
$$n=f_{i_{1}}+f_{i_{2}}+\cdots+f_{i_{k}},\ i_{1}<i_{2}<\cdots<i_{k} .$$
We need to find all $n$ of which representation is unique.
First, if $i_{1} \geq 3$, then we can replace $f_{i_{1}}$ by $f_{i_{1}-1}+f_{i_{1}-2}$ and get another representation of $n$. So if $n$ is a "Fib-unique", $i_{1} \in\{1,2\}$.
Second, if there is some $t$ such that $i_{t+1}-i_{t} \geq 3$, then we can replace $f_{i_{t+1}}$ by $f_{i_{t+1}-1}+f_{i_{t+1}-2}$ and get another representation of $n$.
Hence, for a "Fib-unique" integer $n$, we have $i_{t+1}-i_{t} \leq 2$ for every $0 \leq t \leq k-1$.
Third, if we have some $t$ such that $i_{t+1}-i_{t}=1$, chose that $t$ which is the largest. Then we can replace $f_{i_{t+1}}+f_{i_{t}}$ by $f_{i_{t+1}+1}$, and get another representation of $n$. So for a "fib-unique" $n$ we must have $i_{t+1}-i_{t}=2$ for every $0 \leq t \leq k-1$.
So, every possible "Fib-unique" positive $n$ are $f_{1}, f_{2}$ and those in these forms:
$$\begin{aligned} & \text{ i . } n=f_{1}+f_{3}+\cdots+f_{2k-1}=f_{2k}-1 \\ & \text{ ii . } n=f_{2}+f_{4}+\cdots+f_{2k}=f_{2k+1}-1 \end{aligned}$$
In conclusion, every Fib-unique is $f_{1}, f_{2}$ and $f_{k}-1$ with $k \geq 3$. $\square$