= Solution
A <horseshoe for an interval map> consists of two branches covering a common interval: there are an open interval $J$ and disjoint open subintervals $K_0,K_1\subseteq J$ with $f(K_0)=f(K_1)=J$. Equivalently one can work with two closed intervals with disjoint interiors whose images cover their union, and then restrict the branches. <Glendinning chaos> means that some positive iterate $f^m$ has such a horseshoe, rather than requiring one for $f$ itself.
For the three-cycle, use the <intermediate value theorem> on endpoint images. The <interval covering relations> are $I_a\to I_b$ and $I_b\to I_a,I_b$. Consequently both $f^2(I_a)$ and $f^2(I_b)$ cover $I_a\cup I_b$. This proves \b[$f^2$ has a horseshoe]. The <directed covering graph of an interval map> has exactly these forced arrows; a closed walk $a\to b\to b\to a$ has length three and a primitive itinerary. Pulling back the intervals successively and using the fixed-point property produces a point fixed by $f^3$ with this itinerary. It cannot be fixed by $f$, and a shared endpoint belongs to the already given three-cycle, so a genuine period-three orbit results.
For the increasing spatial order of the five-cycle, let $I_j=[x_{j-1},x_j]$, $1\leq j\leq4$. The forced graph is
$$
I_1\to I_2\to I_3\to I_4,\qquad I_4\to I_1,I_2,I_3,I_4.
$$
The primitive loops $I_4$, $I_3I_4$, $I_2I_3I_4$, and $I_1I_2I_3I_4$ force exact periods $\boxed{1,2,3,4}$. For periods below five the known five-cycle endpoints cannot be responsible, so the interval itineraries give distinct interior orbits. Both $g^2(I_3)$ and $g^2(I_4)$ cover $I_3\cup I_4$, so \b[$g$ is guaranteed to be Glendinning-chaotic].
A horseshoe for $g$ itself is not guaranteed. As a counterexample put $x_j=j$ on $[0,4]$ and interpolate by $g(x)=x+1$ for $0\leq x\leq3$, $g(x)=16-4x$ for $3\leq x\leq4$. This has the required five-cycle. The unique fixed point is $16/5$, with $g(x)>x$ to its left and $g(x)<x$ to its right. Any nondegenerate closed interval whose image covers itself must contain this fixed point in its interior: an interval entirely on either side, or with the fixed point as an endpoint, fails to cover its extreme endpoint. Two intervals with disjoint interiors therefore cannot both cover themselves, as a horseshoe would require. Hence this example has no one-iterate horseshoe.
For the alternative spatial order, set $y_0=x_4,y_1=x_2,y_2=x_1,y_3=x_3,y_4=x_0$ and $J_j=[y_{j-1},y_j]$. Endpoint images give
$$
J_1\to J_4,\quad J_2\to J_2,J_3,\quad J_3\to J_1,\quad J_4\to J_1,J_2.
$$
The self-loop at $J_2$, the two-cycle $J_1J_4$, and the four-cycle $J_2J_3J_1J_4$ force $\boxed{1,2,4}$; period three is not guaranteed. To see this last claim, take the connect-the-dots map at $y_j=j$ with values $(4,3,1,0,2)$. Its exact interval transitions are those displayed. Any length-three closed itinerary is the repeated self-loop in $J_2$; there the map is the affine decreasing function $5-2x$, whose third iterate has only its original fixed point. Endpoints belong to the five-cycle. Thus no exact three-cycle exists in this example.
Chaos is still guaranteed: from $J_2$ there are two different length-four closed walks, four self-loops and $J_2J_3J_1J_4J_2$. Pullbacks give disjoint branches of $g^4$ over a common interval in $J_2$, hence a <horseshoe for an interval map> for an iterate. A horseshoe for $g$ itself is again not guaranteed: the same connect-the-dots example is strictly decreasing on $[0,3]$ and increasing on $[3,4]$ with image $[0,2]$. Its unique fixed point is $5/3$, again with $g(x)>x$ to the left and $g(x)<x$ to the right. The same fixed-point argument rules out two self-covering intervals with disjoint interiors. These counterexamples distinguish the three separate questions about periods, chaos and a one-iterate horseshoe.
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