1)a)
You have to calculate the heat from an electrical resistor  (constant in time).
You have two technicians -
technician A, who can measure voltage to a precision of 10mV
technician B, who can measure amperage (?) to a precision of 0.07*10^(-4) A.

technician A performed 10,000 tests and got V_{average} = 250[mV] with a stddev of 30[mV].
technician A performed 10,000 tests and got I_{average} = 0.2*10^(-4)[A] with a stddev of 0.04*10^(-4)[A].

You only have a budget for one of the following options:

1. buy a new device for technician A, 10 times more accurate, and perform one more test using the new device.

2. fund 10,000 more tests by technician A.

3. buy a new device for technician B, 10 times more accurate, and perform one more test using the new device.

4. fund 10,000 more tests by technician B.

technician B will not reduce its error by performing more tests, because we take as the error max{stddev, instrument error}, and his precision is already higher than stddev.
that isn't the case for technician A. we are left with options 1,2,3.

if we perform one more test with a lower instrument error by technician A, the standard deviation will still be higher than the instrument error, and since we take the error to be max{stddev, instrument error}, this will not lower our error. we are left with options 2,3.

if we perform one more test with an instrument error of 0.007*10^(-4)A, then the stddev will still be almost the same: 0.04*10^(-4)A. since we take the error to be max{stddev, instrument error}, reducing the instrument error will just make us take the stddev, of 0.04*10^(-4)A. it is multiplying the original error by ~0.571.
we will get a relative error of roughly 20%.

we know that error as a function of number of experiments N is error=cons * 1/sqrt(N), so performing another 10,000 experiments for each one, or twice as many, will multiply the stddev by 1/sqrt(2) ~== 0.707.
we will get a relative error of roughly 8.5%

therefore, we should fund 10,000 more tests by technician A to get a lower error.