1. Yes, the difference is constant.

2. It is not really constant, but approximately, yes. With exceptions.

With binary floating point numbers, the expression (f(r)−r)/r is constant within a factor of 2. For rounding error computations, you can assume it is constant.

A note on 0:
Obviously, you cannot apply the formula to r=0. But it is important to know that while f(r)-r decreases when r becomes small, f(0)-0 is much larger than, for instance, f(f(r))-f(r). There is a huge non-representable gap around 0.

1. Yes, the difference is constant.

2. It is not really constant, but approximately, yes. With exceptions.

With binary floating point numbers, the expression $(f(r)−r)/r$ is constant within a factor of $2$. It is between $1 \over 2^m$ and $1 \over 2^{m-1}$ where m is the number of bits in the mantissa. For rounding error calculations, you can assume it is constant.

A note on 0:
Obviously, you cannot apply the formula when $r=0$. But it is important to know that while $f(r)-r$ decreases when $r$ becomes small, $f(0)-0$ is much larger than, for instance, $f(f(r))-f(r)$. There is a huge non-representable gap around $0$.