I think I have an answer, but an not completely sure if it is correct, because first of all it's been a long time I worked with formulaes, and secondly, I have some doubts about how I use Ohm's law.
From what I understand the relationship is T(tip) = T(room) + kP, with:
T = Temperature (Kelvin)
P = power (Watt)
k = constant
Let's define these first:
V = voltage (Volt)
I = current (Amp)
(d) = delta ("the change in")
OK, the change in temperature of the tip is:
T(tip) - T(room) = (d)T = k(d)P
To get rid of that constant, we can say that the change in temperature is proportional to the change in power:
(d)T ~ (d)P
So, if the power is halved, the temperature is also halved. Like in the first question. Now, for the second question:
The power can be calculated using the following formula:
P = V * I
So, you would say that when V is halved, P is also halved, right? No, that's not the case. We can't say anything about P unless we know something about I. And the problem is, we don't know anything about I.
Now comes the trick: Ohm's law states:
V = I * R
which rewrites to:
I = V / R
We can substitute this into the formula for calculating the power:
P = V * (V / R) = V^2 / R
(V^2 means "V square" or V * V)
Now we have a different formula for calculating P. But don't we need to know something about R before we can say something about the change in P? Yes, and we know enough already:
Remember that a soldering iron generates heat becauses of high internal resistance. This value of R is so big, that it changes only little when varying V and I; therefore, this change can be neglected.
(p)P = (p)(V^2) / (p)R
becomes by neglecting the change in R:
(d)P = (d)(V^2) / R
(d)P ~ (d)(V^2)
From this followes that when the voltage is halved, the power is quartered and with it the temperature of the tip.
Note that I'm not sure if I used Ohm's law correctly here. Ohm's law is used for calculating the resistance in an ideal conductor, and of course a soldering iron is far from ideal. However, since we don't actually calculate the resistance here, I think it's save to use it in this way.
Good luck with your exams.