a) We have a well-known theorem: If the function $f(x)$ is continuous on the segment $[a; b]$ then $f(x)$ reaches the maximum value and minimum value at some point on that segment.

Consider the number $f(0) > 0$, because $\lim_{x \to -\infty} f(x) = \lim_{x \to +\infty} f(x) = 0$ then there exist values $a$ small enough and $b$ large enough such that $f(x) < f(0), \forall x \le a$ and $f(x) < f(0), \forall x \ge b$.

Consider the value of $f(x)$ on $[a; b]$, by the above theorem, $f(x)$ reaches maximum value $M = f(c)$ with $M \ge f(0)$ with some $c \in [a; b]$. It is easy to see that for all $x \in (-\infty; a) \cup (b; +\infty)$, $f(x) < f(0) \le M$ so $f(x) \le M, \forall x \in \mathbb{R}$.

Thus, $f(x)$ obtains the maximum value on $\mathbb{R}$.

b) We have the intermediate value theorem: If the function $f(x)$ is continuous and there exist two values $a < b$ such that $f(a)f(b) < 0$ then $f(x) = 0$ has a solution in the interval $(a; b)$.

By this theorem, it is clear that if $f(x)$ reaches two values $A, B$ for some $A < B$ then it reaches all values in the interval $(A, B)$.

Indeed, suppose that $f(u) = A, f(v) = B$ and consider a value $C \in (A; B)$ and the function $g(x) = f(x) - C$. Clearly,
$$g(u) = A - C < 0 \text{ and } g(v) = B - C > 0.$$
It follows that $g(u)g(v) < 0$, so the equation has a solution in $(u, v)$.

Back to the problem, we investigate two following cases:

1) If there exists an interval $(a, b)$ containing $c$ such that $f(c) = M$ and $f(x) < M, \forall x \in (a, b) \setminus \{c\}$.

Take $A, B \neq c$ in the interval $(a, b)$ such that $c \in [A; B]$, from now on we only consider this segment. Since $f(x)$ is continuous on $[A; c]$ and $[c; B]$, there will be a minimum value on these two segments, set as $m_1, m_2$ respectively. Denote $m = \max\{m_1, m_2\}$. According to the intermediate value theorem, there exists $x_1 \in [A; c]$ and $y_1 \in [c; B]$ such that $f(x_1) = f(y_1) = m$; we also have $x_1 < c < y_1$.

Applying this theorem again on segments $[x_1; c]$ and $[c; y_1]$, we see that there exists $x_2, y_2$ such that
$$f(x_2) = f(y_2) = \frac{m + M}{2} \quad \text{and} \quad x_2 < c < y_2.$$
Just like that, we consider the sequence $(u_n)$ such that $u_1 = m$ and $u_{n+1} = \frac{u_n + M}{2}$ for all positive integers $n$. It is easy to see that this sequence converges to $M$ and for every $n \ge 2$, there always exists two numbers $x_n \in [x_{n-1}; c], y_n \in [c; y_{n-1}]$ such that $f(x_n) = f(y_n) = u_n$ and $x_n < c < y_n$.

Note that the sequence $(x_n)$ is increasing and is bounded by $c$, so it has a limit $l \le c$. If $l < c$ then due to continuity, we have $M = \lim u_n = \lim f(x_n) = f(l) < M$, which is a contradiction. Similarly, we have $\lim x_n = c$. So two sequences $(x_n), (y_n)$ are satisfying the problem.

2) If there does not exist an interval $(a, b)$ as above, there will be a segment $[a; b]$ where $f(x) = M, \forall x \in [a; b]$.

We consider the sequence $x_n = \frac{a+b}{2} + \frac{a-b}{2^n}$ and $y_n = \frac{a+b}{2} + \frac{b-a}{2^n}$. It is easy to check that $x_n, y_n \in [a; b]$ for all positive integers $n$, so $f(x_n) = f(y_n) = M$ and
$$x_n < \frac{a+b}{2} < y_n, \lim x_n = \lim y_n = \frac{a+b}{2}.$$
Thus, the problem is solved in all cases. $\square$