= Solution
Each $X_i$ has the same <probability distribution> as $-X_i$. By <independence>, the entire vector and its negative have the same joint law, so $S_m$ has a <symmetric distribution> about zero. The two disjoint tail <events> consequently have equal <probabilities>:
$$
\mathbb P(S_m>m/2)=\mathbb P(S_m<-m/2)
=\frac12\mathbb P(|S_m|>m/2).
$$
Apply <Chebyshev's inequality> with the <mean> and <variance> from (i):
$$
\mathbb P(S_m>m/2)
\le\frac12\frac{2m/3}{(m/2)^2}
=\boxed{\frac4{3m}}.
$$
Using only containment in the two-sided event would give $8/(3m)$; the <symmetry> supplies the necessary extra factor $1/2$.
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