Solution:

(a) $c_{n}=1$ for every $n \geq 2$.
Let us first show that for every $n \geq 2$ one has
$$\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\ldots+\frac{1}{1+a_{n}} \geq 1$$
For $n=2$, setting $a_{1}=a$ we have $a_{2}=\frac{1}{a}$ and the expression on the left-hand side of the inequality reduces to
$$\frac{1}{1+a}+\frac{1}{1+\frac{1}{a}}=\frac{1}{1+a}+\frac{a}{1+a}=1$$
Now suppose $n>2$. The product of the two smallest among $a_{1}, \ldots, a_{n}$ is certainly less than or equal to 1, otherwise at least one of them would be greater than 1 and hence the product $a_{1} a_{2} \cdots a_{n}$ would be greater than 1. Without loss of generality, we may assume that $a_{1} a_{2} \leq 1$. Let $x=\sqrt{\frac{1}{a_{1} a_{2}}}$. Then $x \geq 1$ and $(a_{1} x)(a_{2} x)=1$. We have
$$\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\ldots+\frac{1}{1+a_{n}}>\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}} \geq \frac{1}{1+a_{1} x}+\frac{1}{1+a_{2} x}=1$$
Let us now show that 1 is the largest real number that can be placed on the right-hand side of the inequality, that is, that if $c>1$ the inequality fails for some choice of $a_{1}, \ldots, a_{n}$. Set $a_{1}=a_{2}=\cdots=a_{n-1}=y$, where $y$ is any positive real number such that $y+1>\frac{n-1}{c-1}$, and $a_{n}=y^{-(n-1)}$. We have, for $i=1, \ldots, n-1$,
$$\frac{1}{1+a_{i}}=\frac{1}{1+y}<\frac{c-1}{n-1}$$
and hence
$$\left(\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\ldots+\frac{1}{1+a_{n-1}}\right)+\frac{1}{1+a_{n}}<(n-1) \frac{c-1}{n-1}+1=c .$$

(b) One has $d_{2}=\frac{2}{3}$ and $d_{n}=1$ for every $n \geq 3$.
Let us begin with the case $n=2$. Setting $s=a_{1}+a_{2}$ and taking into account that $a_{1} a_{2}=1$, the term on the left-hand side of the inequality becomes
$$\frac{2+2 s}{5+2 s}=1-\frac{3}{5+2 s}$$
This term attains its minimum value when $s$ is minimal. Since $a_{2}=\frac{1}{a_{1}}$, we have $s=a_{1}+\frac{1}{a_{1}}$ is greater than or equal to 2, and it equals 2 if and only if $a_{1}=1$. Hence the best possible inequality is obtained for $s=2$, giving the value $d_{2}=\frac{2}{3}$.
The rest of the proof is analogous to the previous case. Let us first show that $d_{n} \geq 1$ for every $n \geq 3$. For $n=3$, the inequality
$$\frac{1}{1+2 a_{1}}+\frac{1}{1+2 a_{2}}+\frac{1}{1+2 a_{3}} \geq 1$$
is equivalent to
$$3+4\left(a_{1}+a_{2}+a_{3}\right)+4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right) \geq 1+2\left(a_{1}+a_{2}+a_{3}\right)+4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)+8 a_{1} a_{2} a_{3},$$
that is,
$$a_{1}+a_{2}+a_{3} \geq \frac{8-2}{2}=3$$
which holds by the well-known inequality between arithmetic mean and geometric mean.
Now suppose $n>3$. As before, we may assume that $a_{1} a_{2} a_{3} \leq 1$ and define $x=\sqrt{\frac{1}{a_{1} a_{2} a_{3}}}$. We have $(a_{1} x)(a_{2} x)(a_{3} x)=1$ and hence
$$\begin{aligned} & \frac{1}{1+2 a_{1}}+\frac{1}{1+2 a_{2}}+\frac{1}{1+2 a_{3}}+\ldots+\frac{1}{1+2 a_{n}}>\frac{1}{1+2 a_{1}}+\frac{1}{1+2 a_{2}}+\frac{1}{1+2 a_{3}} \\ & \geq \frac{1}{1+2 a_{1} x}+\frac{1}{1+2 a_{2} x}+\frac{1}{1+2 a_{3} x} \geq 1 . \end{aligned}$$
Finally, if $d$ is any number greater than 1, set $a_{1}=a_{2}=\cdots=a_{n-1}=z$ where $z$ is any positive real number such that $2 z+1>\frac{d-1}{n-1}$, and $a_{n}=z^{-(n-1)}$. We have
$$$$
$\left(\frac{1}{1+2 a_{1}}+\frac{1}{1+2 a_{2}}+\ldots+\frac{1}{1+2 a_{n-1}}\right)+\frac{1}{1+2 a_{n}}<(n-1) \frac{d-1}{n-1}+1=d$ .