Q:

Some simple calculations of current in circuit

I know it's trivial for electronic's person. But I have difficulties with counting following task:

I'll be thankful for any assistance.

A:

In the first circuit R2 and R3 are parallel, so you should be able to calculate their combined value (R23). Once you've done that there's only one path left: the voltage source with R1 and R23 in series, which should enable you to calculate the current and hence the voltage drop over R1 and R23. The voltage over R23 is the voltage over R2 and R3, so you can calculate the currents through each of them. And the current through R1 is just the sum of these.  
The second circuit is more or less the same, but here it's R1 || R3.  
(You get 5 givens, while only 4 are required. The fifth can be derived from the others) 
details 
\$\dfrac{1}{R23} = \dfrac{1}{R2} + \dfrac{1}{R3} = \dfrac{1}{2\Omega} + \dfrac{1}{1\Omega} = \dfrac{1.5}{\Omega} \$.
So \$R23 = \dfrac{1\Omega}{1.5} = 0.67 \Omega\$  
\$I_{E} = \dfrac{E}{R1 + R23} = \dfrac{11V}{3\Omega + 0.67\Omega}=3 A\$  
\$U_{R23} = I_{E} \times R23 = 3 A \times 0.67 \Omega = 2 V\$  
\$I_{R2} = \dfrac{U_{R23}}{R2} = \dfrac{2 V}{2 \Omega} = 1 A\$  
\$I_{R3} = I_{R23} - I_{R2} = 3 A - 1 A = 2 A\$  

