= Solution
Passing to the deterministic-service limit in part (c), near zero, gives
$$
\phi_B(\theta)=\exp\!\left[\frac{\theta+\lambda(\phi_B(\theta)-1)}{\mu}\right].
$$
Differentiation at zero gives $\mathbb EB=(1+\lambda\mathbb EB)/\mu$. Alternatively, the same branching expectation equation gives $\mathbb EB_k=1/(\mu-\lambda)$ for every $k$, because $\mathbb ES_k=1/\mu$. The assumed convergence of finite moment-generating functions at a positive point near zero bounds exponential moments and makes these means uniformly integrable, so the means pass to the limit. Since $T=A+B$, \b[the answers are]
$$
\boxed{\mathbb EB=\frac1{\mu-\lambda},\qquad\mathbb ET=\frac1\lambda+\frac1{\mu-\lambda}}.
$$
They correspond to an <M D 1 queue> with deterministic service time $1/\mu$. The calculation only needs finite transforms near zero; the literal assumption of finite transforms for every $\theta<\lambda$ is stronger than necessary and is not generally true for positive busy-period arguments. For instance, with $k=1$ and $\lambda=\mu/2$, part (c) gives $\lambda b^2-(\mu+\lambda-\theta)b+\mu=0$. Its discriminant is negative for $(\sqrt\mu-\sqrt\lambda)^2<\theta<\lambda$, so no finite busy-period transform can exist there.
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