There are $\binom{18}8$ ways to choose $8$ of the $18$ questions. Let’s count the number that are not acceptable.
There are $\binom{12}8$ ways to choose them so as to have none from the first section, and another $\binom61\binom{12}7$ ways to choose them so as to have exactly $1$ from the first section. Thus, there are $\binom{12}8+\binom61\binom{12}7$ ways to choose them so as to have fewer than $2$ from the first section. There are just as many ways to choose fewer than $2$ from the second section, and just as many again to choose fewer than $2$ from the third section, so as a first approximation there are $3\left(\binom{12}8+\binom61\binom{12}7\right)\tag{1}$ unacceptable choices.
However, $(1)$ is too big, because we counted some unacceptable choices twice. Specifically, any set of $8$ questions that includes fewer than $2$ from two sections got counted as bad for each of those sections, so it got counted twice in $(1)$. To correct for this, we must subtract from $(1)$ the number of unacceptable sets that have too few questions from two sections. Such a set must have all $6$ questions from one section and one question from each of the other two sections. There are $3$ ways to choose the section that contributes $6$ questions and then $6^2$ ways to choose one question from each of the other two sections, so there are $3\cdot6^2$ sets that have fewer than $2$ questions from each of two sections. Thus, $(1)$ should be corrected to $3\left(\binom{12}8+\binom61\binom{12}7\right)-3\cdot6^2\;.\tag{2}$ Since it’s not possible to choose $8$ questions and get fewer than $2$ from each of the three sections, $(2)$ is the number of unacceptable choices. The desired result is therefore $\binom{18}8-3\left(\binom{12}8+\binom61\binom{12}7\right)-3\cdot6^2\;;$ if I’ve done the arithmetic correctly, this is $43~758$.
(In this problem it’s equally easy to count the number of acceptable sets directly, as Viliam Búr has done in his answer.)