Well, speed depends upon frame of reference of course... but if you start in the particle's rest frame, and then attempt to accelerate it, the formula is that E = mc²/(sqrt(1-v²/c²)). This includes the rest mass energy, so when v = 0, you have E = mc², where m is the rest mass. If you do a mathematical expansion (called the Taylor expansion) on this expression, you will get E = mc² + 0.5mv² + (... higher order terms that are very small for v << c). The 0.5 mv² term is the classical kinetic energy that you will learn about in any basic physics class (only valid for speeds v much less than c).

The sqrt of a negative number is not impossible... we have something for it called imaginary numbers defined as i = sqrt (-1). Then you can create the sqrt of any negative number out of it (sqrt (-1)*sqrt (a) = sqrt (-1*a) = sqrt (-a)). In fact, a lot of science (including electrical engineering, signal analysis, and quantum mechanics) depends on imaginary numbers existing, and give very accurate calculations that computers, DVDs, and just about anything electronic depends on. The problem with the tachyon mass being imaginary is that it doesn't make any physical sense to us. Even in quantum mechanics in which a lot of the mathematics depends on imaginary numbers, anything physically measurable (energy, mass, size, etc.) always gives a real (non-imaginary) value. Mass is supposed to be something you can measure, like a book is a 1 kilogram, or a proton is 1.67*10 ^(-27) kilograms, etc. But it doesn't make any sense to us to say that a particle has 1.58 + 3.42*sqrt (-1) kilograms; this is the reason why it's a non-sensible answer.

Dividing by zero is not well defined. In mathematics, division is defined as the inverse of multiplication. In other words, x/y =? really means "what multiplied by y gives x" or if x/y = c, then this literally means "c*y = x." If however y = 0, then the question mark is literally no number. (Anything times zero is zero. Nothing multiplied by zero can give x, for x non-zero. However, 0/0 is not very well defined either, except in limits). However, as you approach calculus, you learn about limits. And in that case, it is sensible to talk about the limit as b goes to zero with some rate, but that's too complicated to discuss without a bit more algebra on your part. But for b = 0, it is not well defined. However, as b->0 (as b approaches zero) with some speed you can talk about how quickly x goes to infinity. And for 0/0, you can take a limit on that as well. When you learn Leibnitz's approach to calculus, you will see what I mean.

Imaginary numbers do work in much the same way as real numbers (that's what they're called. Negative numbers are real, as are fractions, and decimals, non-repeating infinite decimals, etc. In fact, I don't know if you've encountered imaginary numbers yet... it usually isn't mentioned until the first or second year of algebra). The reason why we don't like them in something like the tachyon mass is that we think all things physical have real values (i.e. why we can measure 1kg, but we've never measured something with an imaginary amount of kilograms). That's one of the reasons why we don't like tachyons (there are other people that aren't so realistic, but even so they agree that the theories with imaginary masses are unstable. Therefore even if there's a possibility of something with imaginary mass existing, we would never see it. A simple analogy is that it's like trying to sit a ball completely unmoving on top of another ball. if they're perfect spheres and everything were just perfect with no wind, no interaction with anything else, and they were placed just right on top of each other it might happen for a second, but chances are it would roll down. That's an unstable system. The tachyons are a more complicated of course, but basically the whole theory is unstable).

Well, the reason why people say "the mass is infinite" for a particle with rest mass traveling at the speed of light is because they consider the physical mass you would measure. To make it simpler, I will call the rest mass m still. So if a particle isn't moving, E = mc². Now, let's set it in motion... then E = mc²*1/sqrt (1-v²/c²). But, if we call x = m/sqrt (1-v²/c²), then as v->c we see that x->infinity. Therefore, the energy approaches infinity, and the measured mass x does as well (notice here that it has the same units as mass). And basically that's due to the fact that you're putting a lot of energy (which has a mass due to Einsteins relation) into something. And if you want to make v->c, then you need more and more energy as you get closer and closer to c, but you could never get there.