kirchhoff rules

(1) In accordance to the figure, both terminals of the 7-Ω resistor are connected to the positive terminals of the battery and hence the p.d. across it is zero.
Therefore the current through it is zero.
Then, for the remaining 3-Ω and 4-Ω resistors, they are connected in parallel to the cell and hence both p.d. across them are 10V.
So the current through the 3-Ω resistor = 10/3 A
Current through the 4-Ω resustor = 10/4 = 2.5 A
(2) As follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazyelectric8.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazyelectric9.jpg

1.resolve the circuit you will get this configuration
As the 7ohm resistor is in short-circuit, current through it = 0 A

─10 V ─
│ │
─3ohm ─
│ │
─4ohm ─

Kirchhoff rule states that I out = I in,
in the setup,
method 1,
equvalent resistance = 12/7
so I out = 10/(12/7) = 5.833
I in 3 ohm : 5.833 x 4/7 = 3.333
I in 4 ohm : 5.833 - 3.333 = 2.500

method 2,
both 3 ohm and 4 ohm resistors have 10 V pd,
by V= IR,
I in 3 ohm = 10/3 = 3.333
I in 4 ohm = 10/4 = 2.5

for the second question, the link of image is not availabe.