= Solution
Let $p_{ab}$ denote the conditional winning <probability> in the <CHSH game>. With both settings zero, <computational-basis measurements> of the <Bell state> always agree, so $p_{00}=1$. When just one setting is one, Bob's conditional <pure state> from part (i), or the corresponding calculation with Alice and Bob exchanged, gives agreement <probability> $\cos^2\theta$. Hence $p_{01}=p_{10}=\cos^2\theta$.
For $a=b=1$, winning requires different output <bits>. Alice obtains each of her two outcomes with <probability> $1/2$. Given her first outcome, Bob's desired overlap is
$$
\langle f_1(\theta)|e_0(\theta)\rangle=2\sin\theta\cos\theta=\sin2\theta;
$$
given her second outcome, the other desired overlap is $\langle f_0(\theta)|e_1(\theta)\rangle=-\sin2\theta$. Thus $p_{11}=\sin^2(2\theta)$. Averaging over the four equally likely inputs gives
$$
\boxed{p_{\mathrm{win}}(\theta)=\frac{1+2\cos^2\theta+\sin^2(2\theta)}4
=\frac{3+\cos2\theta-\cos^2(2\theta)}4.}
$$
The maximum winning <probability> for noncommunicating classical players is $3/4$, the game form of the <CHSH inequality>. Setting $t=\cos2\theta$, strict improvement requires $t-t^2>0$, or $0<t<1$. On the stipulated interval this gives
$$
\boxed{0<\theta<\frac\pi4.}
$$
The endpoints give equality with the classical bound. Within this <reflected real-basis CHSH strategy>, the maximum is $13/16$ at $t=1/2$, namely $\theta=\pi/6$; this follows by completing the square in $t$.
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