how to solve this question?

Picture 1:
You can simply the 2 resistors in parallel. The equivalent resistance (only when 2 in parallel) is the product divided by the sum. In your case: (2 * 2) / (2 + 2) = 4/4 = 1 Ω

Picture 2:
You can draw the equivalent resistance.

Picture 3
You can draw the 3 points a, b and c.
Do you know the Kennelly's theorem? Go to this link.
https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_de_Kennelly

Picture 4
You can identify the 3 impedances: Z(at) ; Z(bt) ; Z(ct)
According to the theorem (see the link above), you can write:

Z(at) = Z(ab) * Z(ac) / [Z(ab) + Z(bc) + Z(ac)] = (2 * 1) / [2 + 2 + 1] = 2/5 Ω

Z(bt) = Z(ab) * Z(bc) / [Z(ab) + Z(bc) + Z(ac)] = (2 * 2) / [2 + 2 + 1] = 4/5 Ω

Z(ct) = Z(ac) * Z(bc) / [Z(ab) + Z(bc) + Z(ac)] = (1 * 2) / [2 + 2 + 1] = 2/5 Ω

Picture 5
You can identify the 3 impedances: Z(at) ; Z(bt) ; Z(ct) and the values of them.

Picture 6
You can see a similar drawing.
You can calculate the equivalent resistance adding themselves when they are in serial link.
Branch above: R = 2 + (4/5) = (10/5) + (4/5) = 14/5 Ω
Branch below: R = 1 + (2/5) = (5/5) + (2/5) = 7/5 Ω

Picture 7
You can see a similar drawing.

Picture 8
You can calculate the equivalent resistance: (14/5) Ω in parallel with (7/5) Ω
R = [(14/5) * (7/5)] / [(14/5) + (7/5)]
R = (98/25) / (21/5)
R = 98/105 Ω

Picture 9
You can calculate the equivalent resistance: (14/5) Ω in parallel with (7/5) Ω
R = (98/105) + (2/5)
R = (98/105) + (21/105)
R = (119/105) Ω