= Solution
Let $c$ be the initial call price. Write the risk-neutral <probabilities> of stock prices $5,3,2$ as $q_5,q_3,q_2$. The stock's <martingale> equation and the normalization give $q_2=2q_5$ and $q_3=1-3q_5$. The call pays only in the first state, so its <martingale> equation is $q_5=c$. Thus
$$
(q_5,q_3,q_2)=(c,1-3c,2c).
$$
All three objective <probabilities> are positive. This defines an <equivalent martingale measure> precisely when $0<c<1/3$, and part (iii) proves that every such price excludes <arbitrage>.
To prove necessity without appealing to an existence theorem, consider the other prices explicitly. If $c\leq0$, buy one call and hold $-c$ units of cash. Its initial value is zero and terminal value is $(S_1-4)^+-c$, nonnegative and positive with positive <probability>. If $c\geq1/3$, take holdings $(h_B,h_S,h_C)=(c-1,1/3,-1)$. The initial value is $c-1+1-c=0$ and terminal values at $5,3,2$ are $(c-1/3,c,c-1/3)$. They are nonnegative and the middle-state value is strictly positive. This includes the endpoint $c=1/3$. Consequently
$$
\boxed{\text{the arbitrage-free call prices are exactly }c\in(0,1/3).}
$$
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