Answer
a) $64\;\rm J$
b) $1.6$
Work Step by Step
a) We know, for refrigerator heat engines, that
$$Q_H=W_{in}+Q_C$$
Hence, the heat extracted is given by
$$Q_C=Q_H-W_{in}\tag 1$$
where $Q_H$ is shown in the given graph (105 J), and
$$W_{in}=-W_s=-(78-119)=\bf 41\rm \;J$$
Plugging the known into (1);
$$Q_C=Q_H-W_{in}=105-41$$
$$Q_C=\color{red}{\bf 64 }\;\rm J$$
----
b) The coefficient of performance of a refrigerator is given by
$$K=\dfrac{Q_C}{W_{in}}$$
Plugging the known;
$$K=\dfrac{64}{41}=\color{red}{\bf 1.6}$$
