II. (50 points) There are three types of stocks. The sum of the number of shares of the first two types equals the number of shares of the third type. The total value of the second type of stock is four times that of the first type. The total value of the first and second types of stocks equals the total value of the third type. The price per share of the second type of stock is 1.6 to 2 yuan higher than that of the first type, and the value per share of the third type of stock is no less than 4.2 yuan and no more than 6 yuan. Find the maximum and minimum percentages of the total number of shares that the first type of stock represents.
Let the number of shares of the first, second, and third types of stocks be $k$, $m$, and $n$, respectively, and the value of each share be $x$, $y$, and $z$ (yuan). The problem requires finding the value of:
$$
f = \frac{k}{k+m+n} \times 100\%
$$
From the given information,
$$
\begin{array}{l}
k+m=n, \quad k x+m y=n z, \quad 4 k x=m y. \\
1.6 \leqslant y-x \leqslant 2.0, \quad 4.2 \leqslant z \leqslant 6.0.
\end{array}
$$

Let $p = \frac{m}{k}$. Eliminate $n$ from equation (1) and express $f$ in terms of $p$, and $x$ and $z$ in terms of $y$ and $p$:
$$
\begin{aligned}
f & = \frac{k}{k+m+n} \times 100\% \\
& = \frac{k}{2(k+m)} \times 100\% = \frac{50}{1+p}\%, \\
x & = \frac{p}{4} y, \\
z & = \frac{k x + m y}{k+m} = \frac{x + p y}{1+p} = \frac{\frac{p y}{4} + p y}{1+p} = \frac{5 p y}{4(1+p)}.
\end{aligned}
$$

From equation (2), we get $1.6 \leqslant y - \frac{p y}{4} \leqslant 2.0$,
which simplifies to $1.6 \leqslant \frac{y(4-p)}{4} \leqslant 2.0$,
$$
\begin{array}{l}
\text{Thus, } \frac{6.4}{4-p} \leqslant y \leqslant \frac{8}{4-p}. \\
\text{Also, } \because 4.2 \leqslant \frac{5 p y}{4(1+p)} \leqslant 6.0, \\
\therefore \frac{16.8(1+p)}{5 p} \leqslant y \leqslant \frac{24(1+p)}{5 p}.
\end{array}
$$

For any $a$, $b$, $c$, and $d$, we have
$$
\left\{\begin{array}{l}
a \leqslant y \leqslant b, \\
c \leqslant y \leqslant d
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a \leqslant y \leqslant d, \\
c \leqslant y \leqslant b.
\end{array}\right.\right.
$$

Therefore, equations (3) and (4) hold if and only if
$$
\frac{6.4}{4-p} \leqslant \frac{24(1+p)}{5 p}, \quad \frac{16.8(1+p)}{5 p} \leqslant \frac{8}{4-p}.
$$

Solving these, we get $\frac{7}{3} \leqslant p \leqslant 3$,
hence $12.5\% \leqslant f \leqslant 15\%$.
Therefore, the maximum percentage of the first type of stock in the total of the three types of stocks is $15\%$, and the minimum percentage is $12.5\%$.