1. **Problem statement:** Given parallelogram ABCD, point M lies on the ray opposite to AD such that $3MA=2AD$, and point N lies on segment CM such that $4NC=NM$. We need to express vector $\overrightarrow{AN}$ in terms of vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$. 2. **Set up vectors:** Let $\overrightarrow{AB}=\mathbf{b}$ and $\overrightarrow{AD}=\mathbf{d}$. 3. **Find $\overrightarrow{AM}$:** Since M lies on the ray opposite to AD starting at A, and $3MA=2AD$, vector $\overrightarrow{AM}=-\frac{2}{3}\mathbf{d}$ (because M is on the opposite direction of $\mathbf{d}$). 4. **Find $\overrightarrow{CM}$:** Vector $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=\mathbf{b}+\mathbf{d}$. Then, $$\overrightarrow{CM}=\overrightarrow{AM}-\overrightarrow{AC} = -\frac{2}{3}\mathbf{d} - (\mathbf{b}+\mathbf{d}) = -\mathbf{b} - \frac{5}{3}\mathbf{d}.$$ 5. **Find $\overrightarrow{AN}$:** Point N lies on segment CM such that $4NC=NM$. Let $\lambda$ be the ratio along $\overrightarrow{CM}$ from C to N, so $$\overrightarrow{CN} = \lambda \overrightarrow{CM}.$$ Since $4NC=NM$, and $NM = CN - CM = (\lambda -1)\overrightarrow{CM}$, we have $$4|NC| = |NM| \Rightarrow 4|\lambda \overrightarrow{CM}| = |(\lambda -1)\overrightarrow{CM}| \Rightarrow 4|\lambda| = |\lambda -1|.$$ Because N lies on segment CM, $0<\lambda<1$. Solve: $$4\lambda = 1 - \lambda \Rightarrow 5\lambda =1 \Rightarrow \lambda = \frac{1}{5}.$$ 6. **Calculate $\overrightarrow{AN}$:** $$\overrightarrow{AN} = \overrightarrow{AC} + \overrightarrow{CN} = (\mathbf{b} + \mathbf{d}) + \frac{1}{5}(-\mathbf{b} - \frac{5}{3}\mathbf{d}) = \mathbf{b} + \mathbf{d} - \frac{1}{5}\mathbf{b} - \frac{1}{3}\mathbf{d} = \left(1 - \frac{1}{5}\right)\mathbf{b} + \left(1 - \frac{1}{3}\right)\mathbf{d} = \frac{4}{5}\mathbf{b} + \frac{2}{3}\mathbf{d}.$$ **Final answer:** $$\boxed{\overrightarrow{AN} = \frac{4}{5} \overrightarrow{AB} + \frac{2}{3} \overrightarrow{AD}}$$ This corresponds to option A.