= Solution
The map $F$ itself has a horseshoe only for $\mu=2$. Indeed, its <Lipschitz constant> is $\mu$. If two intervals $J_0,J_1$ form a horseshoe and their convex hull has length $L$, then
$$
L\leq\mu|J_i|\quad(i=0,1),
\qquad
|J_0|+|J_1|\leq L,
$$
which forces $\mu\geq2$. At $\mu=2$, take $J=(-1,1)$, $K_0=(-1,0)$ and $K_1=(0,1)$; both $K_i$ map onto $J$.
The <Lipschitz constant> of $F^2$ is $\mu^2$, so the same length argument makes $\mu\geq\sqrt2$ necessary for an $F^2$ horseshoe. It is sufficient. Put
$$
a=\frac1{1+\mu},\qquad b=\frac{\mu+2}{\mu(1+\mu)}.
$$
For $\mu\geq\sqrt2$, one has $a<1/\mu\leq b\leq1$. With
$$
J=(a,1),\qquad K_0=(a,1/\mu),\qquad K_1=(1/\mu,b),
$$
the two relevant monotone branches satisfy
$$
F^2(a)=F^2(b)=a,
\qquad
F^2(1/\mu)=1.
$$
Thus $K_0,K_1\subset J$ are disjoint and $F^2(K_0)=F^2(K_1)=J$. Therefore
$$
\boxed{F\text{ has a horseshoe iff }\mu=2,\qquad F^2\text{ has one iff }\mu\geq\sqrt2}.
$$
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