= Solution
On upper exit, $b\leq S_T\leq b+c$; on lower exit, $-a-c\leq S_T\leq-a$. With $p=\mathbb P(S_T\geq b)$ and $\mathbb ES_T=0$,
$$
pb-(1-p)(a+c)\leq0
\leq p(b+c)-(1-p)a.
$$
Solving gives
$$
\frac a{a+b+c}\leq p\leq\frac{a+c}{a+b+c}.
$$
Moreover $(S_T+a)(S_T-b)\geq0$, so $\mathbb ES_T^2\geq ab$, which is stronger than the requested lower bound. Also $S_T\in[-a-c,b+c]$, and hence
$$
(S_T+a+c)(b+c-S_T)\geq0.
$$
Taking expectations and using $\mathbb ES_T=0$ gives $\mathbb ES_T^2\leq(a+c)(b+c)$, stronger than the requested upper bound. Since
$$
ab\geq\frac{ab(a+b)}{a+b+c},
\qquad
(a+c)(b+c)\leq
\frac{(a+c)(b+c)(a+b+2c)}{a+b+c},
$$
the two stated estimates follow from part a.
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