Euler could have easily solved it!

Relevant wiki: Euler's Totient Function

Calculating $\phi{(49)}$ :

$\phi{(49)} = 49 \left(1-\dfrac{1}{7}\right) = 42$

Rewriting the problem:

${(3^{48})}^{42}+{(6^{48})}^{42}$

Using Euler's Totient Theorem, $a^{\phi(n)} \equiv 1 \mod n$ :

$1+1 \mod 42$

$\boxed 2$