Why do i get three different methods to calculate the same current ,but get three different current values?
Method 3 is correct.  

\$I_R=I_{R1}+I_{R2}+I_{R3}=\color{orange}{0\text{ A }}+\color{magenta}{[3\text{ A }\frac{(2\Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}]} +\color{blue}{[1\text{ A }\frac{3\Omega}{3\Omega+(2\Omega+1\Omega+R_L)}] } =\frac{18\Omega}{6\Omega+R_L}\text{ A}\$  

That method also shows what goes wrong in method 1 and 2:  

In the Left hand circuit:  

\$I_R=I_1=3 \text{ A } \frac{(2\Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}\$  

The current of 3A isn't divided across \$R_3\$, \$R_2\$, \$R_4\$ and \$R_L\$ like that, because it does not take into account the contribution of the 1A current.

You can see it comparing it to the solution of method 3: Only the contribution of the 3A current source (magenta part) is taken into account by this method 1, not the contribution of the 1A current source (blue part). 

Right hand circuit:  

\$I_R=I_1=3\text{ A }+I_2=3\text{ A }+1\text{ A }\frac{3\Omega}{3\Omega+(2\Omega+1\Omega+R_L)}\$

The problem here is that the 3A current is not only running through \$R_9\$ and \$R_L\$, but also through \$R_8\$ and \$R_7\$.

Comparing to method 3 you can see that only \$ \frac{(2\Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}\$ part of the 3A current flow through \$R_L\$.



Regarding method 1 correct approach would be:

Using the left hand circuit:

The voltage across \$R_2\$ is given by \$V_{R_2}= (i_2+1\text{ A }) \cdot R_2 \$
The voltage across \$R_3\$ is given by \$V_{R_3}=i_2 \cdot  R_3 \$

The sum of these voltages is across \$R_1\$ and \$R_L\$, so the current through \$R_L\$ is given by 
$$ i_{R_L} = \frac{ V_{R_2}+V_{R_3} }{R_1+R_L} = \frac{ (i_2+1\text{ A }) \cdot R_2 + i_2 \cdot  R_3 }{R_1+R_L}$$

According KCL, \$i_2+i_{R_L}=3\text{ A }\$. Substituting for \$i_2\$ this yields 

$$ i_{R_L} = \frac{ (3\text{ A }-i_{R_L} +1\text{ A }) \cdot R_2 + (3\text{ A }-i_{R_L}) \cdot  R_3 }{R_1+R_L}$$

$$ i_{R_L} (R_1+R_L) =  4\text{ A } \cdot R_2 - i_{R_L} \cdot R_2 + 3\text{ A } \cdot  R_3 - i_{R_L}\cdot  R_3 $$

$$ i_{R_L} (R_1+R_2+R_3+R_L) =  4\text{ A } \cdot R_2 + 3\text{ A } \cdot  R_3 $$

$$ i_{R_L} =  \frac{ 4\text{ A } \cdot R_2  + 3\text{ A } \cdot  R_3 }{R_1+R_2+R_3+R_L} $$