Answer
a) $811\;\rm J$
b) $-488\;\rm J$
c) $0\;\rm J$
Work Step by Step
a) From 1 to 2 is an isobaric process, so the heat transferred is given by
$$Q=nC_{\rm p}(T_2-T_1)\tag 1$$
Now we need to find $T_1$ and $T_2$;
$$T_1=\dfrac{P_1V_1}{nR}=\dfrac{(4\times 1.013\times 10^5)(800\times 10^{-6})}{(0.1)(8.31)}=\bf 390\;\rm K$$
$$T_2=\dfrac{P_2V_2}{nR}=\dfrac{(4\times 1.013\times 10^5)(1600\times 10^{-6})}{(0.1)(8.31)}=\bf 780\;\rm K$$
Plugging these two results into (1); recalling that, for monatomic gas, $C_{\rm p}=20.8$
$$Q=(0.1)(20.8)(780-390) =\color{red}{\bf 811}\;\rm J$$
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b) From 2 to 3 is an isochoric process, so the heat transferred is given by
$$Q=nC_{\rm v}(T_3-T_2)\tag 2$$
Now we need to find $T_3$;
$$T_3=\dfrac{P_3V_3}{nR}=\dfrac{(2\times 1.013\times 10^5)(1600\times 10^{-6})}{(0.1)(8.31)}=\bf 390\;\rm K$$
Plugging these two results into (2); recalling that, for monatomic gas, $C_{\rm v}=12.5$
$$Q=(0.1)(12.5)(390-780) =\color{red}{\bf -488}\;\rm J$$
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c) The final temperature is equal to the initial temperature, $T_1=T_3$, so the change in thermal energy of the system is zero.
$$\Delta E_{th}=\color{red}{\bf 0}\;\rm J$$
