$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?

The Carmichael function value for $100$, $\lambda(100)=\text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$ gives us $1234 \equiv 14 \bmod \lambda(100) \implies 3^{1234} \equiv 3^{14} \bmod 100$.
Probably the easiest way to calculate $3^{14} \bmod 100$ is as $81^3 \cdot 9$ discarding higher digits, so $81^2 \equiv 61$ and $81\cdot 61 \equiv 41$ leading to $3^{1234} \equiv 3^{14} \equiv 69 \bmod100$.
Starting alternatively from your observation that $3^{15} \equiv 7 \bmod 100$, we could find $3^{14} \equiv 7\cdot 3^{-1} \equiv 7\cdot 67 \equiv 469 \equiv 69 \bmod 100$.
