Let the two supplement angles be 3x and 7x.
\(\therefore\) 3x+7x = 180°​​​​​​​
[\(\because\) the sum of two supplement angles = 180°​​​​​​​​​​​​​​]
\(\Rightarrow \quad 10x=180°\quad \Rightarrow \quad x=\frac { 180° }{ 10 } \Rightarrow x=18°\)
So, the angles are 3\(\times\)18 = 54°​​​​​​​ and 7\(\times\)18=126°​​​​​​​​​​​​​​

Let x be an angle, whose supplement is \(\frac { 2 }{ 3 } \) of it.
We know that, sum of supplmentary angles is 180°
So, \(x+\frac { 2 }{ 3 } x=180°\)
\(\frac { 3x+2x }{ 3 } =180°\Rightarrow \frac { 5x }{ 3 } =180°\)
\(x=\frac { 180°\times 3 }{ 5 } =36°\times 3=108°\Rightarrow x=180°\)

Yes, \(\angle \)1 is vertically opposite to \(\angle \)4.

No, two obtuse angles cannot form a linear pair because the sum of two obtuse angles is always greater than 180°.

Yes, two adjacent angles can be complementary, if their sum is 90°.
e.g. 
 
\(\because \quad \angle PQS+\angle RQS=90°\)
\(\angle \)PQS and \(\angle \)RQS are two adjacent angles because they have common vertex Q and common arm QS but they do not have a common interior point.

The supplement angle of 125° is 55°.

Let the supplement angle of 90° be x°,
We know that, the sum of two supplementary angles is 180°
\(\therefore\) x°+ 90° = 180° \(\Rightarrow\) x° = 180° - 90° = 90°
Hence, the supplement angle of 90° is 90°.

No, two obtuse angles cannot be supplementary because the measure of an obtuse angle is more than 90°. So, the sum of two obtuse angles would be more than 180°.

135°

Given, a II d, b II e, c II f, a, b, c, d, e and f are line segments.
Give them points A, B, C, 0, E, F, G, H, I, J and K.

\(\because\) \(\angle\)AKE =120°
\(\therefore\)\(\angle\)JKL = \(\angle\)AKE = 120°
[vertically opposite angles)
Now, a ll d
\(\therefore\) \(\angle\)JKL + \(\angle\)KCH=180°          [cointerior angles]
\(\Rightarrow\) 120° + \(\angle\)z = 180°
\(\Rightarrow\) \(\angle\)z = 180° - 120° = 60°
Also, \(\angle\)FGH + \(\angle\)KCH = 180°   [cointerior angles]
\(\Rightarrow\) LFGH = 180° - 60° = 120°
\(\angle\)FGH= \(\angle\)BGD      [vertically opposite angles]
\(\Rightarrow\) \(\angle\)BGD =120° \(\Rightarrow\) \(\angle\)y=120°
and \(\angle\)x + \(\angle\)FGH = 180°  [cointerior angles]
\(\therefore\) \(\angle\)x = 180°-120° = 60°
Hence, \(\angle\)x = 60°, \(\angle\)y = 120° and \(\angle\)z = 60°.

Angle between band c = 30°
[vertically opposite angles]