Find the last $3$ digits of $3^{352}$? Find the last $3$ digits of $3^{352}$ ?
Apart from Carmichael Function, any other way of solving it ?

As you imply, the Carmichael function $\lambda(1000) = \text{lcm}(\,\lambda(5^3), \lambda(2^3)\,) = \text{lcm}(100,2) = 100$ is the least universal exponent.
All$\bmod 1000$, this gives us that $3^{352}\equiv 3^{52}$ and then:
$\begin{align}3^{52} =  9^{26} = (10-1)^{26}  
  & \equiv  (-1)^{26} + 26\cdot 10\cdot (-1)^{25} + \overbrace{\frac{26\cdot25}2}^{\text{odd mult of 5}}\cdot 10^2\cdot (-1)^{24}\\
  & \equiv 1-260+500 \equiv 241\\
\end{align}$
