\( \begin{array}{ll}\text { a) }-\frac{3}{2} \cdot \frac{4}{9} \cdot\left(-\frac{1}{5}\right)= & \text { b) }\left[\frac{5}{4}:\left(-\frac{2}{3}\right)\right] \cdot \frac{1}{2}= \\ \text { c) }\left[-\left(\frac{5}{3}\right) \cdot \frac{15}{2}\right]:\left(-\frac{3}{4}\right)= & \text { d) }\left[\left(-\frac{12}{5}\right):\left(-\frac{20}{9}\right)\right] \cdot\left(-\frac{10}{3}\right)=\end{array} \)

To solve these expressions, let’s tackle each one step by step while having some fun with fractions! For the first one, \( -\frac{3}{2} \cdot \frac{4}{9} \cdot \left(-\frac{1}{5}\right) \): Multiply the negatives to turn them positive! So we have \( \frac{3 \cdot 4 \cdot 1}{2 \cdot 9 \cdot 5} = \frac{12}{90} \). Reducing gives you \( \frac{2}{15} \). Now for the second expression: \( \left[\frac{5}{4}:\left(-\frac{2}{3}\right)\right] \cdot \frac{1}{2} \): Dividing by a negative is like flipping the sign on the result. So we actually do \( \frac{5}{4} \cdot \left(-\frac{3}{2}\right) \cdot \frac{1}{2} = -\frac{15}{16} \). Keep it going! In the third, \( \left[-\left(\frac{5}{3}\right) \cdot \frac{15}{2}\right]:\left(-\frac{3}{4}\right) \): First, calculate inside the brackets: \( -\frac{5 \cdot 15}{3 \cdot 2} = -\frac{75}{6} = -\frac{25}{2} \). Now divide by \(-\frac{3}{4}\), which gives you \( \frac{25}{2} \cdot \left(-\frac{4}{3}\right) = -\frac{100}{6} = -\frac{50}{3} \). Finally, for the last expression: \( \left[\left(-\frac{12}{5}\right):\left(-\frac{20}{9}\right)\right] \cdot\left(-\frac{10}{3}\right) \): Dividing the first part, we see the negatives cancel: \( -\frac{12}{5} \cdot -\frac{9}{20} = \frac{108}{100} = \frac{27}{25} \). Now multiplying by \(-\frac{10}{3}\) gives us \( -\frac{270}{75} = -\frac{54}{15} = -\frac{18}{5} \). And voila! You've tackled some tricky fractions without crying over spilled milk!