a.
Let us denote
$$a_k = \frac{x_k}{1+x_1+\cdots+x_k} \quad \text{and} \quad b_k = 1-a_k = \frac{1+x_1+\cdots+x_{k-1}}{1+x_1+\cdots+x_k}, \quad 1 \le k \le n.$$
By the inequality of arithmetic and geometric means, we have
$$b_1 + b_2 + \cdots + b_n \ge n \sqrt[n]{b_1 b_2 \cdots b_n} = \frac{n}{\sqrt[n]{1+x_1+\cdots+x_n}} = \frac{n}{\sqrt[n]{2}}.$$
Therefore,
$$n \cdot m \le a_1 + a_2 + \cdots + a_n = n - (b_1 + b_2 + \cdots + b_n) \le n \left(1 - \frac{1}{\sqrt[n]{2}}\right),$$
which implies $m \le 1 - \frac{1}{\sqrt[n]{2}}$.

We will show that there exists a sequence $(x_1, x_2, \dots, x_n)$ of positive real numbers summing to 1 such that $a_1 = a_2 = \dots = a_n = 1 - \frac{1}{\sqrt[n]{2}} \stackrel{\text{def}}{=} m_0$, which implies that the greatest possible value of $m$ is $m_0$.
The existence of such a sequence is equivalent to solving the system
$$x_1 + \cdots + x_n = 1, \quad \frac{x_1}{1+x_1} = \frac{x_i}{1+x_1+\cdots+x_i}, \quad \text{for all } i = 2, \dots, n.$$
For $i = 2$, we get $x_2 = a_1(1 + x_1)$. For $i = 3$, replacing $x_2$, we find $x_3 = x_1(1 + x_1)^2$. By induction on $i$, it follows that $x_i = x_1(1 + x_1)^{i-1}$, for all $i \le n$. Since their sum is 1, we get
$$1 = x_1 + x_1(1 + x_1) + \dots + x_1(1 + x_1)^{n-1} = x_1 \frac{(1+x_1)^n - 1}{(1+x_1) - 1} = (1+x_1)^n - 1.$$
Hence, $(1+x_1)^n = 2$, so $x_1 = \sqrt[n]{2}-1$. Substituting back, $x_k = \sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)$, $k=1, 2, \dots, n$.

b.
We will prove that the smallest possible value of $M$ is $M_0 = 1 - \frac{1}{\sqrt[n]{2}}$, which is attained if and only if $a_1 = a_2 = \dots = a_n$.
Using the same reasoning as before, we have $a_1 = a_2 = \dots = a_n$ if and only if $x_k = \sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)$, $k = 1, 2, \dots, n$.
Let $\alpha_k = \sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)$ for $1 \le k \le n$, and consider a positive sequence $(x_1, x_2, \dots, x_n)$, different from $(\alpha_1, \alpha_2, \dots, \alpha_n)$, with sum 1. Then there exists $k \in \{1, \dots, n\}$ such that $\alpha_k < x_k$.
Let $k_0$ be the smallest such index for which $\alpha_{k_0} < x_{k_0}$; then $x_i \le \alpha_i$, for all $1 \le i \le k_0 - 1$. We deduce
$$\begin{aligned} a_{k_0} &= \frac{x_{k_0}}{1 + x_1 + \cdots + x_{k_0}} = \frac{1}{\frac{1+x_1+\cdots+x_{k_0-1}}{x_{k_0}} + 1} > \frac{1}{\frac{1+\alpha_1+\cdots+\alpha_{k_0-1}}{\alpha_{k_0}} + 1} \\ &= \frac{\alpha_{k_0}}{1 + \alpha_1 + \cdots + \alpha_{k_0}} = M_0, \end{aligned}$$
therefore, $M = \max a_k \ge a_{k_0} > M_0$.
In conclusion, $M_0 = 1 - \frac{1}{\sqrt[n]{2}}$ is the minimal possible value of $M$.