Hint for (a): both sequences are monotone and bounded.
Hint for (b): you can explicitely find $\bar b_n$ and $\underline b_n$. Your intuition is correct, btw.
Hint for (c): $\bar b_n\ge\underline b_n$ always.
Hint for (d): $\bar b_n\ge a_n\ge\underline b_n$ to prove in one direction and use the fact that a convergent sequence satisfies the Cauchy criterion to prove in another direction.
Anothe way to understand $\limsup$ and $\liminf$ is to take all convergent subsequences; then find $\sup$ and $\inf$ of the set of their limits. This approach, for example, instantly gives the answer to (b).
Let
$$x^* = \limsup x_n$$
and
$$\alpha_n = \sup_{k\geq n}x_k$$
Since $x_n$ is bounded, $\alpha_n$ is bounded below and non-increasing.
Hence,
$$\lim_{n \rightarrow \infty}\sup_{k\geq n}x_k= \inf_{n}\sup_{k\geq n}x_k=x^*$$
Given $n \in \mathbf{N}$ there exists $m_n \geq n$ such that
$$x^* - 1/n < \alpha_{m_n} < x^* + 1/n.$$
Since $x^*-1/n < \sup_{k\geq m_n}x_k$, there exists $k_n \geq m_n$ such that
$$x^* - 1/n < x_{k_m} \leq\alpha_{m_n}< x^* + 1/n.$$
Hence, $|x_{k_n} - x^*| < 1/n$ where $k_n \geq n$ and the subsequence $(x_{k_n}$) converges to $x^*$.
You can make a similar argument for $\liminf x_n$.
To show that no subsequence can converge to a value greater than $\limsup x_n$, assume that a subsequence converges to $x' > x^*$. Let $\epsilon = (x'-x^*)/2$. Then there are infinitely many $x_n$ greater than $x^* + \epsilon$, a contradiction of the basic property of $x^* = \limsup x_n$.
Again, try to make similar argument for $\liminf x_n$.
Best Answer
The presentation is just a little sloppy, in that the authors have not previously explained that a sequence that increases (decreases, resp.) without bound is said to converge to $+\infty$ ($-\infty$, resp.), but the idea should be clear anyway. It is not really contradictory: it is just an extension of the earlier definition.
And it is entirely compatible with that definition once you know a bit of topology. One forms the extended real numbers by adjoining points $-\infty$ and $+\infty$ and giving them nbhds analogous to $\epsilon$-nbhds $(a-\epsilon,a+\epsilon)$ around real numbers. Specifically, just as we say that a sequence $\langle x_n:n\in\Bbb N\rangle$ converges to $a$ if and only if for each $\epsilon>0$ there is an $n_0\in\Bbb N$ such that $x_n\in(a-\epsilon,a+\epsilon)$ for all $n\ge n_0$. we say that it converges to $+\infty$ if and only if for each $b\in\Bbb R$ there is an $n_0\in\Bbb N$ such that $x_n>b$ for all $n\ge n_0$. The open rays
$$(b,\to)=\{+\infty\}\cup\{x\in\Bbb R:x>b\}$$
function as the equivalent at $+\infty$ of open intervals $(a-\epsilon,a+\epsilon)$ at $a$. And just as at $a$ it generally suffices to work just with the countable family of intervals $\left(a-\frac1n,a+\frac1n\right)$, so at $+\infty$ we can generally work just with the countable family of rays $(m,\to)$ for $m\in\Bbb Z$.
The notion of convergence of a sequence to $-\infty$ is defined similarly: $\langle x_n:n\in\Bbb N\rangle$ converges to $-\infty$ if and only if for each $b\in\Bbb R$ there is an $n_0\in\Bbb N$ such that $x_n<b$ for all $n\ge n_0$.
From a topological point of view convergence to $\infty$ or $-\infty$ behaves no differently from convergence to some $a\in\Bbb R$; both are instances of a more general notion of sequential convergence. And it is convenient to make the extension even at this early point, because sequences that diverge in $\Bbb R$ because they converge to $\infty$ or $-\infty$ in the extended reals, rather than for other reasons (like $\langle (-1)^n:n\in\Bbb N\rangle$, for instance) are quite well-behaved and can often be treated right along with those that converge to real numbers.