This is an answer to your "actual question" (2), building on some of the ideas in Douglas Zare's answer.
Lemma 1: Suppose that $0 < r < 1$. Let $S=\lbrace \epsilon r^i : \epsilon = \pm 1 \text{ and } i \in \mathbb{Z}_{\ge 0} \rbrace$. Fix $k \ge 1$. Let $S_k$ be the set of sums of the form $s_1+\cdots+s_k$ such that $s_i \in S$ and $|s_1|=1$ and there is no nonempty subset $I \subset \lbrace 1,\ldots,k \rbrace$ with $\sum_{i \in I} s_i = 0$. Then $0$ is not in the closure of $S_k$.
Proof: Use induction on $k$. The base case is trivial: $S_1=\lbrace -1,1\rbrace$. Now suppose $k \ge 2$. If a sequence $(x_i)$ in $S_k$ converges to $0$, then the smallest summand in the sum giving $x_i$ must tend to $0$, since a lower bound on the absolute values of the summands rules out all but finitely many elements of $S_k$, which are all nonzero. Discarding the finitely many $x_i$ for which the smallest summand is $\pm 1$ and removing the smallest summand from each remaining $x_i$ yields a sequence $(y_i)$ in $S_{k-1}$ tending to $0$, contradicting the inductive hypothesis.
Now fix $b>1$ and $k$. Let $T=\lbrace \epsilon \lfloor b^n + 1/2 \rfloor : \epsilon = \pm 1 \text{ and } n \in \mathbb{Z}_{\ge 0} \rbrace$. Let $T_k$ be the set of sums of the form $t_1+\cdots+t_k$ with $t_i \in T$.
Lemma 2: Each $t=t_1+\cdots+t_k \in T_k$ equals $u_1+\cdots+u_\ell+\delta$ for some $\ell \le k$ and some $u_i \in T$ with $u_i = O(t)$ and $\delta = O(1)$.
Proof: Examine the powers of $b$ used in the $t_i$. If any nonempty subsum (with signs) of these powers equals $0$, the corresponding $t_i$ sum to $O(1)$. If $b^n$ is the largest power that remains after removing all such subsums, divide all the remaining $t_i$ by $b^n$, and apply Lemma 1 with $r=1/b$ to see that $|t|/b^n$ is bounded away from $0$, so all these remaining $t_i$, which are $O(b^n)$, are $O(t)$.
Corollary: The number of elements of $T_k$ of absolute value less than $B$ is $O((\log B)^k)$ as $B \to \infty$.
Corollary: $T_k \ne \mathbb{Z}$.
This is not a solution, just some thoughts which are too long for a comment. I have added proofs of Will Jagy and Junkies comments/conjectures which are fairly interesting on their own.
First, the Fibonacci numbers are a divisibility sequence, which means that $$\gcd(F_n,F_m)=F_{\gcd(n,m)}.$$
Added: Proof of part of Will Jagy's Observation:
Claim: If $n\equiv 5\pmod{6}$ then $L_n$, and hence $F_{2n}$ are divisible by some prime $p\equiv 3 \pmod{4}$.
Proof: Look at $L_n$ modulo $4$. Then the sequence is $L(0)\equiv 2$, $L(1)\equiv 1$, $L(2)\equiv 3$, $L(3)\equiv 0$, $L(4)\equiv 3$, $L(5)\equiv 3$, $L(6)\equiv 2$, $L(7)\equiv 1$, and at this point it must repeat. The cycle length is $6$, and $L(5)\equiv 3$. This means that $L(5+6k)\equiv 3\pmod{4}$ for all $k$. Hence $L(5+6k)$ is always divisible by a prime congruent to $3$ mod $4$.
Since $L_n |L_{kn}$ when $k$ is odd, we can conclude that if $p\equiv 5 \pmod{6}$ divides $n$, then some prime $q\equiv 3\pmod{4}$ must divide $F_{2n}$. This is because either $2|n$, and hence $3|F_{2n}$, or $n$ is odd, and $L_p|L_n$ so that $q|F_{2n}$.
Added Proof Of Junkie's Comment:
Claim: The density of even Fibonacci numbers which are not divisible by some prime of the form $3+4k$ is $0$.
Proof: This is a corollary of Will Jagy's observation. Since the density of numbers which are not divisible by a prime of the form $5+6k$ is zero, it follows from the previous claim that the density of even Fibonacci numbers not divisible by a prime of the form $3+4k$ is $0$.
Conjectures and other thoughts: Recall that we can write $F_n$ as a sum of two squares if it has no prime factors of the form $3+4k$.
Conjecture 1: The only Fibonacci number of the form $F_{2n}$ which is divisible by some prime of the form $3+4k$ and can be written as the sum of two squares is $F_{12}$.
$F_{12}$ is a very special Fibonacci number for a few reasons. One is that it is the only nontrivial square. If we change the condition to a sum of two nonzero squares, then $F_{12}$ is automatically excluded. Also, since no other Fibonacci numbers are squares, nothing else is affected. Hence we can rephrase conjecture 1 as:
Conjecture 1: If $F_{2n}$ is divisible by some prime of the form $3+4k$ then it cannot be written as the sum of two nonzero squares.
One reason for this conjecture is that it checks out numerically in large range, up to $F_{1000}$. Also, it would be nice if it was true.
Assuming this, by the divisibility property, given a prime $p=3+4k$, we need only care about the first time it appears in the Fibonacci Sequence. There is a theorem which states that $$F_{p-\left(\frac{p}{5}\right)}\equiv 0 \pmod{p},\ \ \ \ \ \ \ \ \ (1) $$ where $\left(\frac{p}{5}\right)$ is the Legendre symbol. Conjecture 1 would imply that if $2n$ is divisible by $p-\left(\frac{p}{5}\right)$ for any $p\equiv 3 \pmod{4}$, then $F_{2n}$ is not the sum of two nonzero squares.
Previously, I said some things about what happens if the above were an "if and only if" for primes of the form $3+4k$ (it clearly isn't for $1+4k$) Small update: It also is just false for $3+4k$, since $3571=3+4k$, is prime and divides $F_{68}$.
Examples: The first few primes congruent to $3$ mod $4$ less then $100$ are $$3,7,11,19,23,31,41,47,59,67,71,79,83,87$$
and they divide respectively
$$F_4, F_8, F_{10}, F_{18}, F_{24}, F_{30}, F_{40}, F_{48}, F_{58}, F_{68}, F_{70}, F_{78}, F_{84}, F_{88}.$$
So in particular, (assuming conjecture 1) if I want $F_{2n}$ to be a sum of two nonzero squares, $2n$ cannot be divisible by any of the above. I.e. we cannot have any of $$2|n, \ 5|n, \ 9|n,\ 29|n,\ 39|n .$$ This idea can give us a large list of primes, none of which can divide $n$, but that is about all I can get it to do.
Best Answer
It follows from the identity
$$F_{n+k} = \sum_{j=0}^k {k \choose j} F_{n-j}$$
which is obtained by applying the standard recurrence $k$ times to the left side, each time splitting up each term into two terms.
Indeed, this gives
$$\sum_{k=0}^{\infty} \frac{F_{n+k}}{k!} = \sum_{k=0}^{\infty} \frac{ \sum_{j=0}^k {k \choose j} F_{n-j}}{k!}= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{j! (k-j)!} F_{n-j} = \sum_{j=0}^{\infty} \sum_{k-j=0}^{\infty} \frac{1}{(k-j)!} \frac{F_{n-j}}{j!} = \sum_{j=0}^{\infty} e \frac{F_{n-j}}{j!}$$