Evaluating $a^n \mathrm{~mod~}m$ is easy when $m$ is prime or $a, m$ are co-prime meaning $\gcd(a,m)=1$. But when $a, m$ are not co-prime, it's not easy to solve it. In this case, Repeated Squaring method can resolve it which needs just about $\lceil\log_2{n}\rceil$ steps at most. Although here, $3$ and $79$ are co-prime, we're going to use repeated squaring method as it's the generalized one.
$3^{37} \mathrm{~mod~} 79\equiv 3\left(3^{18}\right)^2\mathrm{~mod~} 79 \tag{1}$
$3^{18} \mathrm{~mod~} 79\equiv \left(3^{9}\right)^2\mathrm{~mod~} 79 \tag{2}$
$3^{9} \mathrm{~mod~} 79\equiv 3\left(3^{4}\right)^2\mathrm{~mod~} 79 \tag{3}$
$3^{4} \mathrm{~mod~} 79\equiv 81\mathrm{~mod~} 79 \tag{4}$
Now evaluating in a bottom-up fashion,
no$(4)$ $\Rightarrow 3^{4} \equiv 2$
no$(3)$ $\Rightarrow 3^{9} \equiv 3\left(2\right)^2\equiv3\times4\equiv12$
no$(2)$ $\Rightarrow 3^{18} \equiv \left(12\right)^2\equiv144\equiv 65\equiv-14$
no$(1)$ $\Rightarrow 3^{37} \equiv 3\left(-14\right)^2\equiv3\times196\equiv3\times38\equiv114\equiv35$
$\therefore 3^{37} \mathrm{~mod~} 79\equiv35$
So the correct answer is D.
Note: Here $\equiv$ means modular equivalence.