Power Source = I3 x 12 VPower Source = 7.38 W

[Mosaness]

1. Making appropriate use of resistor combination techniques, calculate i3 in the circuit of Fig. 3.84 and the power provided to the circuit by the single current source.



2. Connected in Parallel and Series. Ohm's Law. KCL/KVL



3. I honestly am just guessing how to do this one:

The 3 ohms and 5 ohms resistors are in parallel to one another, so combining their resistance should give 1.875 ohms. The 3 ohms and the 9 ohms resistors too are in parallel, so they should give a total resistance of 2.25 ohms, which can be added to the 3 ohms and 5 ohms in the rightmost branch, to give 10.25 ohms. That leaves us with another pair of resistors in parallel, the 3 ohms and the 6 ohms, which when combined give us 2 ohms.

Now we have two resistors in parallel. The resistor of 1.875 ohms and the resistor of 12.25 ohms. This gives a total resistance of 1.626 ohms.

This can be used to find what I3 is:

1 - 1.626(I3) = 0

I3 = 0.615 A

Now that we know what I3 is, we can calculate the power source using the current.

 

[Bhumble]

Assuming that you're analysis of resistors is correct (seemed like you had the right approach but I didn't double check), the equation you stated isn't correct because this is a current source not a voltage and your equation is mixing the two.
This problem is to make sure you understand KCL.
I_s = I3 +I2 + I1 = v/R_t = v/(R3+R2+R1)
I_s = 1A and voltage is the same to all branches.

I'd start with finding the equivalent resistance branch(es) and the total resistance. Calculate voltage then go about finding branch currents.

 

[ehild]

1. Making appropriate use of resistor combination techniques, calculate i3 in the circuit of Fig. 3.84 and the power provided to the circuit by the single current source.



2. Connected in Parallel and Series. Ohm's Law. KCL/KVL



3. I honestly am just guessing how to do this one:

The 3 ohms and 5 ohms resistors are in parallel to one another, so combining their resistance should give 1.875 ohms. The 3 ohms and the 9 ohms resistors too are in parallel, so they should give a total resistance of 2.25 ohms, which can be added to the 3 ohms and 5 ohms in the rightmost branch, to give 10.25 ohms. That leaves us with another pair of resistors in parallel, the 3 ohms and the 6 ohms, which when combined give us 2 ohms.

Now we have two resistors in parallel. The resistor of 1.875 ohms and the resistor of 12.25 ohms. This gives a total resistance of 1.626 ohms.

This can be used to find what I3 is:

1 - 1.626(I3) = 0

I3 = 0.615 A

Now that we know what I3 is, we can calculate the power source using the current.

Your equation (in red) says that 1A = 1.626 (Ω) * I₃ (A) , which is wrong.
The voltage is Vx=3i₃. You have 1-i₃ current flowing through the parallel resultant of the 12.25 Ω resistor and 5 Ω resistor, Rp. Rp(1-i₃)=Vx.

ehild