You are well on the way to the solution of your first question. As you have guessed, you have two pieces of the puzzle: how much energy the star is producing; and how much hydrogen fuel it has.

Therefore you need only know the efficiency of energy production from the fusion of hydrogen (0.007 of the mass-energy equivalent, as you stated). You have to be careful of units here. You will have to put the mass of the star in kilograms (the sun is 1.98x10^30 kg) and multiply by the speed of light in m/s (3 x 10^8 m/s) squared to get the energy equivalent of the star's mass in Joules (watt-seconds). Multiply this further by the efficiency of hydrogen fusion (0.007) to find out how much energy the star can produce from this method. Now take the luminosity of the sun in watts (Joule/s), [you will have to look this up], multiply by 10,000, and then divide this into the energy you got in the previous paragraph to find the time over which this energy rate can be sustained (the answer will be in seconds). Convert to years and you're done.

There is one simplification assumed here -- that all the hydrogen in the star burns. For stars 15 times the mass of the sun, actually only about half of the hydrogen can circulate through the center of the star's core where it burns. The rest stays pretty much put as an inert outer shell. So the age you compute is an upper limit -- the true hydrogen-burning lifetime will actually be about half as long.

As for your second question, you need to refer to the book's explanation of the Planck radiation curve. In essence, this is a graph of how much radiation of a given wavelength is emitted by an object of a given temperature. The formula for the curve itself is complicated, but it has one redeemingly simple property: the peak of the curve occurs at a wavelength which is inversely proportional to temperature. At the temperature of the Sun (roughly 6000K), the peak wavelength of the Planck curve is at green light (roughly 5500 angstroms or 550 nanometers). If you go to 100,000 K, you've increased the temperature by a factor of almost 17, and the peak wavelength (where the majority of energy of the object is radiated) will shift down in wavelength by a corresponding factor. Therefore the radiation of a star at this temperature will be found chiefly at about 330 angstroms (33 nanometers). This is in the far ultraviolet. Because light of wavelength shorter than about 3000 angstroms doesn't penetrate the Earth's atmosphere, such a telescope would need to be in Earth orbit. Indeed, light of wavelength shorter than 912 angstroms doesn't even easily penetrate the hydrogen clouds of interstellar space, and therefore the object would have to be along a line of sight relatively free of such clouds to be perceptible at 330 angstroms. But it would still be profitable to look for it in the not so far ultraviolet as opposed to regular visible wavelengths.