It looks like a ratio problem to me.
Question 5.
Those answers can't be right. From Kirchhoff's law of currents the sum of the currents in those three branches must be 7A.
By inspection the current ratios are:
A: 4
BC: 2
DEFG: 1
Sum of ratios = 7, 7A/7 = 1, so 1 is now the multiplier.
A: 4 x 1 = 4A
BC: 2 x 1 = 2A
DEFG: 1 x 1 = 1A
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Question 6 first network.
The current through the BCD set is 6A with current ratios:
B: 2
CD: 1
Sum of ratios = 3, 6A/3 = 2, so 2 is now the multiplier.
B: 2 x 2 = 4A
CD: 1 x 2 = 2A
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Question 6 second network.
The current through the EFGHIJ set is 6A with current ratios:
EF: 1
GH: 1
IJ: 1
Sum of ratios = 3, 6A/3 = 2, so 2 is now the multiplier.
EF: 1 x 2 = 2A
GH: 1 x 2 = 2A
IJ: 1 x 2 = 2A
A is obviously 6A.
Hope this helps.