Think of the number next to 1999

Relevant wiki: Euler's Theorem

Since 2 and 100 are not coprime, we have to consider $2^{1999} \mod 4$ and $2^{1999} \mod 25$ separately.

$\begin{cases} 2^{1999} \equiv 2\cdot 4^{999} \equiv 0 \pmod 4 \\ 2^{1999} \equiv 2^{1999 \mod \phi (25)} \equiv 2^{1999 \mod 20} \equiv 2^{19} \equiv 1024 \cdot 512 \equiv -12 \equiv 88 \pmod {25} \end{cases}$

Since $88 \equiv \begin{cases} 0 \pmod 4 \\ 88 \pmod {25} \end{cases} \implies 2^{1999} \equiv 88 \pmod {100}$

Now, consider ( edited as suggested by @Alex G ):

$\begin{aligned} x & \equiv 3^{1999} \pmod{100} \\ 3x & \equiv 3^{2000} \pmod{100} \\ & \equiv 3^{2000 \mod \phi(100)} \pmod{100} \\ & \equiv 3^{2000 \mod 40} \pmod{100} \\ & \equiv 1 \pmod{100} \\ \implies 3x & \equiv 201 \pmod{100} \\ \implies x & \equiv 67 \pmod{100} \\ 3^{1999} & \equiv 67 \pmod{100} \end{aligned}$

$\implies 2^{1999}+3^{1999} \equiv 88 + 67 \equiv 155 \equiv \boxed{55} \pmod {100}$

Did same but also calculated by elementary method by writting2^1999 =16^499*2 and using cyclity of two digits of 16 and similarly by using cyclity of 81

Observe that $2^{100}\equiv 76 \pmod{100}$ and $3^{100}\equiv 1 \pmod{100}$ . Thus,

$2^{1999}+3^{1999}\equiv 2^{99}+3^{99}\equiv 88+67 \equiv 55\pmod{100}$