Solution:

a. We call increasing the monotone numbers whose digits appear in increasing order, and similarly we call decreasing the monotone numbers whose digits appear in decreasing order. We further call twin of a monotone number $N$ the integer obtained by replacing every digit of $N$ with its complement to $10$. Thus, for example, the twin of $12578$ is $98532$.

Note that every decreasing number is the twin of an increasing number, and vice versa. Moreover, it is easy to see that the sum of a five-digit monotone number and its twin is always equal to $111110$.

Suppose now that we have written the sum $S$ of all five-digit monotone numbers. In this sum we can group the addends in pairs, so that to each increasing number corresponds its decreasing twin. By the previous observation, performing the sums in pairs, we get that $S$ becomes the sum of addends all equal to $111110$. Moreover, the addends are as many as the increasing numbers, which in turn are as many as the ways to choose five different digits from a set of nine, that is
$$\binom{9}{5} = \frac{9!}{5!\,4!} = 126$$
Without using the combinatorial meaning of binomial coefficients one can reason as follows. First we choose the $5$ different digits that make up the number: the first can be chosen in $9$ different ways, the second in $8$, ..., the fifth in $5$. In doing so we do not always obtain an increasing number: indeed, in order to obtain one, the first digit must be the smallest of the $5$ chosen, which happens only once out of $5$, the second must be the smallest of the remaining $4$, which happens only once out of $4$, and so on. Consequently the increasing numbers are
$$\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{5 \cdot 4 \cdot 3 \cdot 2} = 126$$
Therefore the sum of all five-digit monotone numbers is $126 \cdot 111110 = 13999860$.

b. Let us denote by $M$ the least common multiple of all monotone numbers. We first observe that the set of monotone numbers is finite (because no monotone number can have more than nine digits), hence $M$ is well defined.

To know with how many zeros $M$ ends, we need to know the largest power of $10$ that divides $M$. To this end we consider separately the powers of $2$ and of $5$. Let us begin by analyzing the case of $5$. The largest power of $5$ that divides $M$ equals the largest power of $5$ by which a monotone number is divisible. This power is at least $3$, since $5^{3} = 125$ is a monotone number. Moreover all integers divisible by $125$ necessarily end with $000$, $125$, $250$, $375$, $500$, $625$, $750$, $875$, and consequently the monotone multiples of $125$ can only end with $125$ or $875$. The only monotone numbers having such an ending are $125$, $875$, $9875$, none of which is divisible by $5^{4}$. Therefore the largest power of $5$ that divides $M$ is $5^{3}$.

Since $M$ obviously contains more than three factors of $2$ (for example $2^{4} = 16$ is a monotone number), it follows that $M$ ends with exactly $3$ zeros.