How to find the remainder of $\frac{2^{99}}{9}$? The question is solved in the book I read in a very odd way as,

$$\frac{2^{99}}{9} = \dfrac{{(2^{3})}^{33}}{2^3-(-1)}$$
  Hence by remainder theorem the remainder is $-1$. In questions when remainder is negative than the number is subtracted from numbers like $2^3$ to get a positive number so the answer is $8-{-1}=9$.

I don't know which remainder theorem the book is talking of. I converted the question in mod notation as,
$$2^{99} \bmod 9 = (2^3)^{33} \bmod (2^3 - (-1))$$
But I don't think there is some formula for $a \bmod (c-d)$. Please tell me how to solve these type of questions.

To solve questions like this, the usual approach is using 
$$\def\mod{\mathbin{\rm mod}} ab \mod n = (a \mod n)(b\mod n) \mod n $$
iteratively. Let's start. We have that 
$$ 2^{99} = 2^{4\cdot 24 + 3} = 8 \cdot 16^{24} $$
reducing modulo $9$, we have 
$$ 2^{99}\mod 9 = 8 \cdot 7^{24} \mod 9 $$
But $7^{24} = 49^{12}$, reducing again and continue in that way
\begin{align*}
  2^{99} \mod 9 &= 49^{12}\cdot 8 \mod 9\\
                &= 4^{12} \cdot 8 \mod 9\\
                &= 16^6 \cdot 8 \mod 9\\
                &= 7^6 \cdot 8 \mod 9\\
                &= 49^3 \cdot 8 \mod 9\\
                &= 4^3 \cdot 8 \mod 9\\
                &= 16^2 \cdot 2 \mod 9\\
                &= 49 \cdot 2 \mod 9\\
                &= 8.
\end{align*}
