= Solution
For $1<s\leq\sqrt2$, use <centered tent-map period-doubling renormalization>. Let
$$
c=s-1,\qquad J_s=[-c,c],\qquad h_s(y)=-cy.
$$
The image $T_s(J_s)=[1+s-s^2,1]$ lies to the right of $J_s$, with disjoint interiors, because $1+s-s^2\geq s-1$. It lies in the positive branch. Consequently,
$$
T_s^2(x)=1-s+s^2|x|=-c+s^2|x|\quad(x\in J_s),
$$
and its range $[-c,(s^2-1)c]$ is contained in $J_s$. The orientation-reversing <affine map> $h_s$ gives the exact <topological conjugacy>
$$
\boxed{h_s^{-1}\circ T_s^2\circ h_s(y)=1-s^2|y|=T_{s^2}(y).}
$$
Thus renormalization squares the slope and halves the return period. Since $s>1$, there is a least $n\geq0$ for which $s^{2^n}>\sqrt2$. Minimality implies $s^{2^n}\leq2$, and all previous slopes are in the renormalizable range. Successive <affine maps> therefore conjugate the restriction of $T_s^{2^n}$ on a nested central interval to $T_{s^{2^n}}$.
By the preceding part, an iterate $T_{s^{2^n}}^m$ has a <horseshoe for an interval map>. Pull its two branch intervals and its range interval back through the composed conjugacy. They give a horseshoe for $T_s^{2^n m}$. Therefore
$$
\boxed{T_s\text{ has an iterate with a horseshoe for every }1<s\leq2.}
$$
At the endpoint $s=\sqrt2$, one renormalization gives the full slope-two tent map, so the argument includes that endpoint without assuming the strict inequality from part (a).
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