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: $T_k \ne \mathbb{Z}$.
Pi has some large partial quotients (that's why 22/7 is such a good approximation), your other irrationals don't. Try some other irrational with a real good rational approximation, and let us know what happens.
Best Answer
This maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula $$ F_n = (A^n-B^n)/(A-B), $$ where $A$ and $B$ are the roots of $T^2-T-1$. So if $\sqrt5$ is in your finite field, then so are $A$ and $B$, and since $AB=-1$, the period divides $p-1$ from Fermat's little theorem. If not, then you're in the subgroup of $\mathbb F_{p^2}$ consisting of elements of norm $\pm1$, so the period divides $2(p+1)$. If you want small period, then take primes that divide $A^n-1$, or really its norm, so take primes dividing $(A^n-1)(B^n-1)$, where $A$ and $B$ are $\frac12(1\pm\sqrt5)$. An open question is in the other direction: Are there infinitely many $p\equiv\pm1\pmod5$ such that the period is maximal, i.e., equal to $p-1$?
BTW, the source you quote isn't quite correct, if $p\equiv\pm2\pmod5$, then the period divides $2(p+1)$, but might not divide $p+1$. The simplest example is $p=3$, where the Fibonacci sequence is $$ 0,1,1,2,0,2,2,1,\quad 0,1,1,2,0,2,2,1,\ldots. $$ Note that the first 0 does not necessarily mean that it will start to repeat. What happens is that the term before the first $0$ is $p-1$, so the first part of the sequence repeats with negative signs before you get to a consecutive $0$ and $1$.