Don't understand power relationship circuit analysis problem
I think the biggest issue with this question is that its phrasing is a bit convoluted and easy to misunderstand. Also I very much disagree with calling S2 &quot;open&quot; when S2 is always choosing between one of two possible paths for current flow. Here is how I would rephrase this question:

Part 1: When S2 is connected through a short circuit, how will the power in R1 change when S1 is closed? Will the power in R1 increase, decrease, or stay the same?
Part 2: When S2 is connected through R3, how will the power in R1 change when S1 is closed? Will the power in R1 increase, decrease, or stay the same?

The best way to solve this is with a voltage divider as @thece suggests. As you note, the problem can be analyzed using either \$P = I\cdot V\$ or \$P = {V^2 \over R}\$, so let's choose the approach that saves the step of calculating the current -- let's use \$P = {V^2 \over R}\$.
In Part 1, each side of R1 is connected to V1 via a short circuit. The power in R1 is always \$P = {V_1^2 \over R_1}\$ regardless of whether S1 is open or closed. It's constant.
Part 2 is the trickier case. The voltage across R1 will be determined by a voltage divider. When S1 is open we have a voltage divider of R3 and R1. When S2 is closed we have a voltage divider of R3 and (R1//R2). In either case the power across R1 is determined by the voltage across this voltage divider. Let's call this output voltage \$V_x\$. Whether the switch is open or closed, the power across R1 will be \$P = {V_x^2 \over R_1}\$. But the two cases give us two different values for \$V_x\$.
Switch open:
$$
V_x = V_1 \cdot {R_1 \over {R_1 + R_3}}
$$
Switch closed:
$$
V_x = V_1 \cdot {{R_1 // R_2} \over {R_1 // R_2 + R_3}}
$$
Recall that \$(R_1 // R_2) &lt; R_1\$ and the rest should make sense.