Inspired By Kalpok Guha

You got lucky this time . The first line is flawed. You can apply euler's theorem only when gcd (a,n) =1. But gcd(12,40) is not 1 . 12^40 = 76 mod 100 instead.

Since it has almost been a month and the OP hasn't corrected his solution, I'm giving the solution in the comments here.

We compute the residue of $12^{12^{12}}$ modulo $25$ and $4$ separately and then combine them later using CRT.

First, note that $12\equiv 0\pmod{4}$ . As such, any arbitrary positive exponentiation of $12$ will be congruent to $0$ modulo $4$ . For modulo $25$ , we can use Euler's theorem since $\gcd(25,12)=1$ . We have $\phi(25)=20$ and we get,

$12^{12^{12}}\equiv0\pmod4\\~\\ \begin{aligned}12^{12^{12}}\equiv12^{12^{12}~\bmod~20}\equiv 12^{4^{6}~\bmod~20}&\equiv 12^{(-4)^3~\bmod~20}\\ &\equiv 12^{-4}\\ &\equiv 19^{-2}\\ &\equiv 11^{-1}\equiv 16\pmod{25}\end{aligned}$

The modular inverse of $11$ modulo $25$ is found using Extended Euclidean Algorithm as $\gcd(11,25)=1=\color{#3D99F6}{16}\times (11)+(-7)\times 25$ . So,

$12^{12^{12}}\equiv\begin{cases}0\pmod4\\ 16\pmod{25}\end{cases}\implies 12^{12^{12}}\equiv 16\pmod{100}$

The last step is done using CRT, i.e., the Chinese Remainder Theorem.