Answer
a) $1.49\;\rm m/s^2,\;0\;m/s^2$
b) $-0.275\;\rm m/s^2,\;0\;m/s^2$
Work Step by Step
a) First of all, we need to analyze the 3-N force into its $x$ and $y$ components.
From the geometry of the given figure, their components are
$$F_{x }=3\sin20^\circ{\;\rm N},\;\;\;\;\;\;\;\;\;F_{y}=3\cos20^\circ{\;\rm N}$$
We know, from Newton's second law, that
$$\sum F_{x }=ma_x,\;\;\;\;\;\;\;\;\;\sum F_{y}=ma_y$$
Thus,
$$a_x=\dfrac{\sum F_{x }}{m}=\dfrac{5-1-3\sin20^\circ}{2}=\color{red}{\bf1.49}\;\rm m/s^2$$
And,
$$a_y=\dfrac{\sum F_{y }}{m}=\dfrac{2.82-3\cos20^\circ}{2}=\color{red}{\bf 0}\;\rm m/s^2$$
b) We need to analyze the $\bf vertical$ 2-N force into its $x$ and $y$ components.
From the geometry of the given figure, their components are
$$F_{x }=2\sin15^\circ{\;\rm N},\;\;\;\;\;\;\;\;\;F_{y}=2\cos 15^\circ{\;\rm N}$$
Now we need to analyze the $\bf horizontal$ 2-N force into its $x$ and $y$ components.
From the geometry of the given figure, their components are
$$F_{x }=2\cos15^\circ{\;\rm N},\;\;\;\;\;\;\;\;\;F_{y}=2\sin 15^\circ{\;\rm N}$$
Thus,
$$a_x=\dfrac{\sum F_{x }}{m}=\dfrac{2\cos 15^\circ+2\sin15^\circ -3}{2}=\color{red}{\bf -0.275}\;\rm m/s^2$$
And,
$$a_y=\dfrac{\sum F_{y }}{m}=\dfrac{1.414+2\sin15^\circ- 2\cos15^\circ}{2}=\color{red}{\bf 0}\;\rm m/s^2$$
