Solution:

Let $ABCD$ be the trapezoid, with $AB$ the longer base, and let $P$ and $Q$ be the endpoints of the cut already made, lying respectively on $AD$ and $BC$. Let the oblique sides be extended until they meet at a point which we call $E$; the triangles $DCE$, $PQE$, $ABE$ are similar (they all have congruent angles thanks to the parallelism of the lines $DC$, $PQ$, $AB$). We know that $PQ / DC = 3$ and $AB / DC = 4$, from which, denoting by $S$ the measure of the area of triangle $DCE$, we obtain the ratios of areas: $\operatorname{Area}(PQE) / S = 3^{2} = 9$, $\operatorname{Area}(ABE) / S = 4^{2} = 16$.

It follows that, by subtraction, $\operatorname{Area}(PQCD) = 8S$ and $\operatorname{Area}(ABQP) = 7S$. Suppose we divide the trapezoid $PQCD$ into $m$ parts, and the trapezoid $ABPQ$ into $n$ parts; in order for all the parts to have equal area, it must be that $8S / m = 7S / n$, that is $8n = 7m$, and therefore $m$ must be a multiple of $8$ and $n$ a multiple of $7$. At minimum $m+n$ must equal $7+8=15$.

To divide the initial trapezoid into $15$ parts we must make $14$ cuts (also counting the cut of length $3$ made at the beginning); let us call $P_{i}$ and $Q_{i}$ (respectively on $AD$ and $BC$) the endpoints of the $i$-th cut (in order of length: in this way $P$ will coincide with $P_{8}$ and $Q$ with $Q_{8}$). The area of the trapezoid $P_{i}Q_{i}CD$ equals $i$ times the area of a single part, which is equal to $S$, so that $\operatorname{Area}(P_{i}Q_{i}E) = (i+1)S$. As before, the triangles $P_{i}Q_{i}E$ and $DCE$ are similar; the ratio between corresponding sides equals the square root of the ratio between the areas (which is $i+1$); therefore $P_{i}Q_{i} = \sqrt{i+1}$. In conclusion, the cuts have lengths $\sqrt{2}, \sqrt{3}, \ldots, \sqrt{15}$, and allow the initial trapezoid to be divided into exactly $15$ parts.