= Solution
Replace each implication by a <logical disjunction> and use the <conjunctive normal form> for <exclusive or> from part (a). The resulting <Boolean clauses> are
$$
\begin{array}{c|l}
F_1&C_1=\neg H\lor Y\\
F_2&C_2=\neg T\lor\neg I\\
F_3&C_3=H\lor T,\quad C_4=\neg H\lor\neg T\\
F_4&C_5=Y\lor I,\quad C_6=\neg Y\lor\neg I.
\end{array}
$$
Here the displayed array is a list of the six <Boolean clauses>, whose <logical conjunction> is equivalent to all four constraints.
Use <resolution> first on $I$, then on $T$, and finally on $H$:
$$
\begin{aligned}
C_7&=\operatorname{Res}_{I}(C_2,C_5)=\neg T\lor Y,\\
C_8&=\operatorname{Res}_{T}(C_3,C_7)=H\lor Y,\\
C_9&=\operatorname{Res}_{H}(C_1,C_8)=Y\lor Y=\boxed{Y.}
\end{aligned}
$$
Removing the repeated <Boolean literal> in the final <Boolean clause> uses idempotence. Neither $C_4$ nor $C_6$ is needed: even the weaker assumptions that the coin has at least one outcome and that at least one person wins already force $Y$, once the two implications hold.
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