Answer
a) Equal.
b) Greater than.
Work Step by Step
a) We know, for an ideal gas, that
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$
So if the gas starts from point 1 and ends at point 2, no matter what the process is, then the temperature at point 1 $T_1$ is constant and the temperature at point 2 $T_2$ is also constant.
Thus, the two processes' temperature differences are the same
$$\boxed{\Delta T_A=\Delta T_B}$$
b) Since the temperature change is the same in both cases, the thermal energy change is also the same in both cases.
$$\Delta E_{th,A}=\Delta E_{th,B}$$
Hence,
$$Q_A+W_A=Q_B+W_B$$
We know that the work done on the gas is given by the area under the curve. $$W=-\rm Area\;under\;the \;curve$$
And we can see that the area under the curve A is greater than that under B. Thus, $ W_A \gt W_B $, and hence the work done on A is more negative than the work done on B, which means that $W_B-W_A \gt 0$
Thus,
$$W_B-W_A=Q_A-Q_B\gt 0$$
So,
$$Q_A-Q_B\gt 0$$
$$\boxed{Q_A\gt Q_B}$$
