= Solution
First the exit time has sufficiently many finite moments. Over any interval of length $a^2$, a <Brownian motion> increment has probability $q=\mathbb P(|B_{a^2}|>2a)>0$ of absolute size larger than $2a$. If the process is still inside $(-a,a)$, such an increment forces it to exit before the next interval endpoint. <Independence> of increments at deterministic times therefore gives
$$
\mathbb P(T>ka^2)\leq(1-q)^k.
$$
Thus $T<\infty$ <almost surely> and all its positive moments are finite. This justifies extracting moments from its transform rather than presupposing them.
For $\lambda>0$, put $\theta=\sqrt{2\lambda}$. The process
$$
M_t=e^{-\lambda t}\cosh(\theta B_t)
=\frac12\left(e^{\theta B_t-\theta^2t/2}+e^{-\theta B_t-\theta^2t/2}\right)
$$
is a <martingale>, being the average of two <exponential martingales for Brownian motion>. Apply the <optional stopping theorem> at the bounded time $T\wedge t$. Since $|B_{T\wedge t}|\leq a$, the stopped values are bounded by $\cosh(\theta a)$. Continuity gives $B_T\in\{-a,a\}$. Letting $t\to\infty$ by the <dominated convergence theorem> yields
$$
1=\mathbb E[e^{-\lambda T}\cosh(\theta B_T)]
=\cosh(a\sqrt{2\lambda})\,\mathbb Ee^{-\lambda T}.
$$
Hence the <Laplace transform of symmetric Brownian interval-exit time> is
$$
\boxed{\phi_T(\lambda)=\frac1{\cosh(a\sqrt{2\lambda})},\qquad\lambda\geq0.}
$$
The value at zero follows from almost-sure finiteness. Expanding the reciprocal gives
$$
\phi_T(\lambda)=1-a^2\lambda+\frac56a^4\lambda^2+O(\lambda^3).
$$
Compare with part (a), now applicable because of the geometric tail bound. The <Brownian symmetric interval-exit moments> are
$$
\boxed{\mathbb ET=a^2,\qquad\mathbb ET^2=\frac53a^4,\qquad
\operatorname{Var}(T)=\frac23a^4.}
$$
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