$(2^{3n} - 1)$ will be divisible by $(\forall n \in N)$

The remainder when $((64)^{(64)})^{(64)}$ is divided by $7$ is equal to

The remainder when $5^{99}$ is divided by $13$ is

If $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$,where $k, t \in Z^{+}$,then $(p, q) =$

Given below are two statements:
Statement $I$: $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by $7$.
Statement $II$: The integral part of $(7 + 4\sqrt{3})^{25}$ is an odd number.
In the light of the above statements, choose the correct answer from the options given below:

The expression $2^{4n} - 15n - 1$, where $n \in N$ (the set of natural numbers), is divisible by