how does the first resistor's resistance influences the second
resistor's resistance.
It doesn't. However, the resistance of the first resistor influences the voltage across the second resistor.
Clearly, the resistors in the diagram are series connected thus the current through each resistor is identical.
$$I_1 = I_2 = I$$
By Ohm's law, we have
$$V_1 = I_1 \cdot R_1 = I \cdot R_1$$
$$V_2 = I_2 \cdot R_2 = I \cdot R_2$$
By KVL, we have
$$6V = V_1 + V_2 = I \cdot R_1 + I \cdot R_2 = I \cdot (R_1 + R_2)$$
Thus
$$I = \frac{6V}{R_1 + R_2}$$
In other words, the current \$I\$ depends on the sum of the resistances (series connected resistances add).
The voltage across the second resistor can now be written as
$$V_2 = I \cdot R_2 = \frac{6V}{R_1 + R_2} \cdot R_2 = 6V \frac{R_2}{R_1 + R_2} $$
and so, as first stated, the resistance of the first resistor influences the voltage across the second resistor.
Similarly
$$V_1 = I \cdot R_1 = \frac{6V}{R_1 + R_2} \cdot R_1 = 6V \frac{R_1}{R_1 + R_2}$$
You can't improve the expected accuracy by combining resistors with the same accuracy. But you will improve the standard deviation of the result. What this means is that if you had 10 individual 1 ohm, 5% resistors, the standard deviation of the 10 resistors would be close to the standard deviation of the lot they were manufactured from. However, if you combine 9 of them as you did to form a 1 ohm resistor, and did this 10 times with different resistors from the same lot, the standard deviation of these ten 1 ohm resistor networks would be smaller than the first set by a factor of 3 (the square root of 9). What this means is that you have improved the probability of being closer to 1 ohm by using a network of 9 resistors because you have narrowed the width of the probability distribution around 1 ohm.
Best Answer
Since they share the same current, the end-points cannot tell the difference in an equivalent circuit.
Only a real voltage drop on each and power dissipation may be different for power heat rise.