This is really two separate questions, so I'll answer them that way. I'm also going to rewrite your problem as:

sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8)

Now, we know sqrt(1) and sqrt(4) are 1 and 2 respectively. And sqrt(2) is, well, sqrt(2). We can't really do anything about this one. So we've got:

sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8) = 1 + sqrt(2) + 2 + sqrt(8) = 3 + sqrt(2) + sqrt(8)

What about sqrt(8)? Well, 8 = 2*4 = 2*(2^2). So

sqrt(8) = sqrt[2*(2^2)] = sqrt(2)*sqrt[(2^2)]

But sqrt[(2^2)] = 2 since the square root and the ^2 undo each other. (Or you could simply say sqrt[(2^2)] = sqrt(4) = 2.) So

sqrt(8) = sqrt[2*(2^2)] = sqrt(2)*sqrt[(2^2)] = sqrt(2)*2 = 2*sqrt(2)

So

sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8) = 3 + sqrt(2) + sqrt(8) = 3 + sqrt(2) + 2*sqrt(2)

= 3 + 3*sqrt(2)

-Dan