If magnet of magnetic moment M makes an angle \(\theta \) with the field, then other magnet of magnetic moment M\(\sqrt { 3 } \) makes an angle \(\left( { 90 }^{ \circ }-\theta \right) \) with the field.
In equilabrium, \(\tau _{ 1 }=\tau _{ 2 }\)
\(\Rightarrow MBsin\theta =M\sqrt { 3 } Bcos\theta \ \Rightarrow \ \frac { sin\theta }{ cos\theta } \sqrt { 3 }\) 
\( \\ \Rightarrow tan\theta =\sqrt { 3 } \Rightarrow \ \theta ={ 60 }^{ \circ }\)

(i)  

The direction of magnetic field is from S to N-pole of magnet.
(ii) 

The direction of magnetic field is from N to S-pole of magnet.

here, I = 5 A, dl = 1 cm = 0.01 m, r = 1 m, \(\theta ={ 45 }^{ 0 }\)
  [\(\because \) direction is North-East]
\(\therefore \ dB=\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { Idl \ sin\theta }{ { r }^{ 2 } } \)
\(={ 10 }^{ -7 }\times\cfrac { 5\times0.01\times sin\quad { 45 }^{ 0 } }{ ({ 1) }^{ 2 } } \)
= 3.54 x 10^-9 T
Its direction is vertically downwards.

Frequency of alternating electric field cyclotron is given by
 \(f=\frac { qB }{ 2\pi m } \)
Here, q = 3.2 x 10^-19 C,
 m = 6.7 x 10^-27 kg and B = \(\frac { \pi }{ 2 } T\)
 \(\\ \therefore \quad f=\frac { q(\frac { \pi }{ 2 } ) }{ 2\pi m }\)
\( \\ f=\frac { (3.2\times { 10 }^{ -19 })\times (\frac { \pi }{ 2 } }{ 2\times \pi \times 6.7\times { 10 }^{ -27 } }\)
 f = 1.2 x 10⁷ cycle/s

Here, first we have to find the direction of magnetic field at point O₂ due to the wire carrying current I₁. Use Maxwell's right hand grip (cork screw) rule, the direction of magnetic field at point O₂ due to current I₁ is along Y-axis.
Here,the wire at point O₂ is placed along  Y- axis. Now, by the formula,
F = I₂ (I x B)
Angle between I and B is 0^o ,both are at Y-axis, i.e
 F = IlB sin 0^o = 0
So,the force exerted at point O₂ because of wire along X-axis is zero.