a) Observe that $a_n > 0$ and that $a_{n+1} - 2 = \frac{2-a_n}{a_n}$ for $n \ge 1$, so $a_{2n-1} > 2 > a_{2n}$ for $n \ge 1$. Thus $a_{2n-1}+a_{2n}-4 = \frac{a_{2n-1}^2-3a_{2n-1}+2}{a_{2n-1}} > 0$ for $n \ge 1$.

$|a_n - 2| = \frac{|a_{n-1} - 2|}{a_{n-1}} = \dots = \frac{|a_1 - 2|}{a_{n-1} \dots a_1} \le \frac{1}{2^{[n/2]}} |a_1 - 2| \to 0$ shows that $a_n \to 2$.

b) For $x = 0$ we get $a < \frac{a_1 + \dots + a_n}{n}$, so $a \le \lim_{n \to \infty} \frac{a_1 + \dots + a_n}{n} = 2$. We will show that the inequality is true for $a = 2$ giving $a_{\max} = 2$. It is enough to show that $\sqrt{x^2 + a_{2n-1}^2} + \sqrt{x^2 + a_{2n}^2} > 2\sqrt{x^2 + 4}$, for any $n \in \mathbb{N}^*$, which is equivalent to $a_{2n-1}^2 + a_{2n}^2 + 2\sqrt{(x^2 + a_{2n-1}^2)(x^2 + a_{2n}^2)} > 2x^2 + 16$. As $a_{2n-1} + a_{2n} > 4$ involves $a_{2n-1}^2 + a_{2n}^2 > 8$, it is sufficient to show that $a_{2n-1}a_{2n} \ge 4$. This is an immediate consequence of $a_{2n-1} > 2$ and $a_{2n} = 1 + 2/a_{2n-1}$.