Step 1: Find the total charge passed.
Charge equals current times time, $Q = I\,t$. With $I = 38.6\;A$ and $t = 100\;s$, \[ Q = 38.6 \times 100 = 3860\;C. \]
Step 2: Convert charge into moles of electrons.
One mole of electrons carries $96500\;C$, so \[ n_e = \frac{3860}{96500} = 0.04\;mol. \]
Step 3: Find copper deposited at the cathode.
Copper is reduced by $Cu^{2+} + 2e^- \rightarrow Cu$, so two moles of electrons give one mole of copper. Moles of $Cu = \dfrac{0.04}{2} = 0.02\;mol$.
Step 4: Convert copper moles to mass.
Using molar mass $63.54\;g\;mol^{-1}$, \[ \text{mass} = 0.02 \times 63.54 = 1.27\;g. \]
Step 5: Find the gas liberated at the anode.
With platinum electrodes in $CuSO_4$ solution, water is oxidised at the anode giving oxygen: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. Four moles of electrons release one mole of $O_2$, so moles of $O_2 = \dfrac{0.04}{4} = 0.01\;mol$.
Step 6: Convert oxygen moles to volume at STP.
One mole of gas occupies $22.4\;L$ at STP, so volume $= 0.01 \times 22.4 = 0.224\;L$. The copper consumed and the gas volume are therefore
\[ \boxed{1.27\;g,\;0.224\;L} \]