One
The wording in the first question is not very good. I think they mean that if R1 =5 and r2 = 9, then in circuit B, the values of the two resistors are the same: 5 and 9.
Here's the catch. When you have a parallel circuit, the parallel resistance is smaller than the smallest resistor.
Take any 2 resistors as examples. They are randomly chosen. Say R1 = 3 and R2 = 6.
1/Rt = 1/3 + 1/6
1/Rt = 2/6 + 1/6
1/Rt = 3/6
1/Rt = 1/2 Take the reciprocal of 1/2
Rt = 2
When a voltage is put across these two resistors, the voltage thinks it is seeing a resistance of 2 ohms which is smaller than either 3 or 6.
What does that mean? It means that the current is going to be greater in the combined parallel circuit than it will be in the series circuit.
We're still not ready to declare which answer is correct. This question is quite tricky.
R1 + R2 in series will have a smaller current than in the parallel pair no matter what the value of R1 and R2 are.
That's because the formula is I = E/(R1 + R2)
R1 + R2 > R1*R2/(R1 + R2)
But that means that E/(R1 + R2) < than either E/R1 or E/R2
Now we are ready to declare an answer.
Answer: Circuit B [The parallel circuit]
Two
This problem is so much easier than the first one. In this question, the current is the same in all series components. A series circuit gives a voltage drop, not a change in current.
The answer is A 2.5 amperes in all 3 resistors.