Answer
See below.
Work Step by Step
Proofs using mathematical induction consist of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number. Then we prove that then the statement also holds for $n + 1$.
Hence, here we have:
1) For $n=1: F_{3(1)}=2$, this is even.
2) Assume for $n=k: F_{3k}$ is even. Then for $n=k+1$:
$F_{3(k+1)}=F_{3k+3}=F_{3k+2}+F_{3k+1}=F_{3k+1}+F_{3k}+F_{3k+1}=2F_{3k+1}+F_{3k}$. The first term of the sum is even because of the factor $2$ and the second term of the sum is even because of the inductive hypothesis. Thus, their sum is even as well.
Thus we proved what we wanted to.