II. (40 points) Let the number of ways to choose $k$ pairwise coprime numbers from the set $\{1,2, \cdots, 28\}$ be $T(k)$. Find $T(2)+T(3)+\cdots+T(12)$.
Obviously, the prime numbers in $\{1,2, \cdots, 28\}$ are 2, 3, $5,7,11,13,17,19,23$, a total of 9.

When $k \geqslant 11$, the $k$ numbers taken out, except for 1, must include at least 10 numbers. By the pigeonhole principle, there must be two numbers divisible by the same prime number, which means these two numbers are not coprime, leading to a contradiction.
Therefore, $T(k)=0(k \geqslant 11)$.
Thus, $T(2)+T(3)+\cdots+T(12)$ is the number of all subsets $M$ of $\{1,2, \cdots, 28\}$, where $|M| \geqslant 2$, and the numbers in $M$ are pairwise coprime.

Below, we calculate $T$ by considering the prime factorization of the numbers in $M$.
(1) All numbers in $M$ have exactly one prime factor, then $M$ is formed by taking one number from each of the following sets:
$$
\begin{array}{l}
\left\{2,2^{2}, 2^{3}, 2^{4}\right\},\left\{3,3^{2}, 3^{3}\right\},\left\{5,5^{2}\right\},\{7\}, \\
\{11\},\{13\},\{17\},\{19\},\{23\}
\end{array}
$$

Similar to the calculation of the number of divisors of a number, the number of $M$ is
$$
T_{1}=(1+4)(1+3)(1+2)(1+1)^{6}-1=3839 \text {, }
$$
where there are 15 single-element subsets.
(2) $M$ has exactly one number $a$ with two prime factors, and the rest of the numbers have only one prime factor.
The prime factors of $a$ are $\{2,3\}, \{2,5\}, \{2,7\}, \{2,11\}$, $\{2,13\}, \{3,5\}$, or $\{3,7\}$. Then $a$ is one of $6i (i=1,2,3,4)$, $10i (i=1,2)$, $14i (i=1,2)$, $22$, $26$, $15$, or $21$, and the rest of the numbers are taken from the remaining sets in (1). The number of $M$ is $T_{2}=3616$, where there are 12 single-element subsets.
(3) $M$ has exactly two numbers $a, b$ each with two prime factors, and the rest of the numbers have only one prime factor.
The prime factors of $a, b$ are $\{2,5\}$ and $\{3,7\}$, $\{2,7\}$ and $\{3,5\}$, $\{2,11\}$ and $\{3,5\}$, $\{2,11\}$ and $\{3,7\}$, $\{2,13\}$ and $\{3,5\}$, $\{2,13\}$ and $\{3,7\}$. Then $(a, b)$ is one of $(10i, 21) (i=1,2)$, $(14i, 15) (i=1,2)$, $(22,15)$, $(22,21)$, $(26,15)$, $(26,21)$,

The rest of the numbers are taken from the remaining sets in (1). The number of $M$ is $T_{3}=288$.
(4) $M$ has three numbers $a, b, c$ each with two prime factors, let these six prime numbers be $p_{i} (i=1,2, \cdots, 6)$. Then
$$
p_{1} \geqslant 2, p_{2} \geqslant 3, p_{3} \geqslant 5, p_{4} \geqslant 7, p_{5} \geqslant 11, p_{6} \geqslant 13 \text {. }
$$

Since $13 \times 3 > 28$, $p_{6}$ can only be a prime factor of 2, but $p_{5} \times 3 \geqslant 11 \times 3 > 28$, leading to a contradiction.
(5) $M$ has one number $a$ with three prime factors, then $a \geqslant 2 \times 3 \times 5 > 28$, leading to a contradiction.

In summary, and considering the inclusion of "1" and the single-element subsets in (1) and (2), we have
$$
T=2\left(T_{1}+T_{2}+T_{3}\right)-27=15459 .
$$