Let the first figure be \( x \) and the second figure be \( y \).
From the problem, we can derive two equations:
1. Quarter of the first figure is equal to three times the second figure:
\[
\frac{x}{4} = 3y
\]
Therefore, multiplying both sides by 4:
\[
x = 12y \quad \text{(Equation 1)}
\]
2. If 12 is added to four times the second figure, the resultant figure equals three times the first figure:
\[
4y + 12 = 3x
\]
Substituting \( x \) from Equation 1:
\[
4y + 12 = 3(12y)
\]
Simplifying this gives:
\[
4y + 12 = 36y
\]
Rearranging the equation gives:
\[
12 = 36y - 4y
\]
\[
12 = 32y
\]
Dividing by 32:
\[
y = \frac{12}{32} = \frac{3}{8}
\]
Now, substituting \( y \) back into Equation 1 to find \( x \):
\[
x = 12y = 12 \times \frac{3}{8} = \frac{36}{8} = \frac{9}{2} = 4.5
\]
Thus, the first figure \( x \) is \( 4.5 \) and the second figure \( y \) is \( \frac{3}{8} \).
To summarize:
- First figure \( x = 4.5 \)
- Second figure \( y = \frac{3}{8} \)
In sequence: \( 4.5, \frac{3}{8} \).