Solution:
(a) Let us assume that there are no even numbers in the sequence. This means that $x_{n+1}$ is obtained from $x_{n}$, by adding a nonzero even digit of $x_{n}$ to $x_{n}$, for each $n \geq 1$.
Let $E$ be the left most even digit in $x_{1}$ which may be taken in the form
$$x_{1} = O_{1} O_{2} \cdots O_{k} E D_{1} D_{2} \cdots D_{l}$$
where $O_{1}, O_{2}, \ldots, O_{k}$ are odd digits $(k \geq 0)$; $D_{1}, D_{2}, \ldots, D_{l-1}$ are even or odd; and $D_{l}$ odd, $l \geq 1$.
Since each time we are adding at least 2 to a term of the sequence to get the next term, at some stage, we will have a term of the form
$$x_{r} = O_{1} O_{2} \cdots O_{k} E 999 \cdots 9 F$$
where $F = 3, 5, 7$ or $9$. Now we are forced to add $E$ to $x_{r}$ to get $x_{r+1}$, as it is the only even digit available. After at most four steps of addition, we see that some next term is of the form
$$x_{s} = O_{1} O_{2} \cdots O_{k} G 000 \cdots M$$
where $G$ replaces $E$ of $x_{r}$, $G = E+1$, $M = 1, 3, 5$, or $7$. But $x_{s}$ has no nonzero even digit contradicting our assumption. Hence the sequence has some even number as its term.

(b) If there are only finitely many even terms and $x_{t}$ is the last term, then the sequence $\langle x_{n}\rangle_{n=t+1}^{\infty} = \langle x_{t+1}, x_{t+2}, \ldots \rangle$ is obtained in a similar manner and hence must have an even term by (a), a contradiction. Thus $\langle x_{n}\rangle_{n=1}^{\infty}$ has infinitely many even terms.