Answer
See the detailed answer below.
Work Step by Step
a) For monatomic gas the change in thermal energy is given by
$$\Delta E_{th}=nC_{\rm v}\Delta T$$
Hence the change in temperature is given by
$$\Delta T=\dfrac{\Delta E_{th}}{nC_{\rm v}}\tag 1$$
And since the thermal energy of the gas is increased by 1 J, so the change in the thermal energy is this q J.
Plugging the known
$$\Delta T=\dfrac{(1)}{(1)(12.5)}=\color{red}{\bf 0.08}\;\rm K=\color{red}{\bf 0.08}^\circ C$$
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b) For diatomic gas, we can use (1) again;
$$\Delta T=\dfrac{\Delta E_{th}}{nC_{\rm v}} $$
Plugging the known
$$\Delta T=\dfrac{(1)}{(1)(20.8)}=\color{red}{\bf 0.048}\;\rm K=\color{red}{\bf 0.048}^\circ C$$
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c) For solids, we can use (1) again;
$$\Delta T=\dfrac{(1)}{(1)(25)}=\color{red}{\bf 0.04}\;\rm K=\color{red}{\bf 0.04}^\circ C$$
