Last two digits

$N=\displaystyle\prod_{k=0}^{166} 7+12k$

We will consider $N$ mod $4$ and mod $25$ separately

$\begin{aligned} N &=\displaystyle\prod_{k=0}^{166} \mathbin{\color{#D61F06} 7}+\mathbin{\color{#3D99F6} 12k}\\ & \equiv \displaystyle\prod_{k=0}^{166} \mathbin{\color{#D61F06} -1} + \mathbin{\color{#3D99F6} 0}\\ & \equiv (-1)^{167} \equiv -1 \equiv 3 \mod 4 \end{aligned}$

The product contains the term $7+12\cdot 14$ and we have

$7+12\cdot 14=7(1+12\cdot 2)=7\cdot 25\equiv 0 \mod 25$

Which implies $N\equiv 0 \mod 25$

We now have the following system of congruences

$\begin{cases} N\equiv 3 \mod 4\\ N\equiv 0 \mod 25 \end{cases}$

And by the CRT we have only one solution mod $100$ , namely $N\equiv \boxed{75}\mod 100$