Answer:
(a)
Oxidation
Na -> Na+ + e
Reduction
Cu2+ + 2e -> Cu
(b) Just multiply oxidation half-reaction by 2 and reduction half-reaction by 1.
2Na -> 2Na+ + 2e
Cu2+ + 2e -> Cu
(c) 2Na + Cu2+ -> 2Na+ + Cu (2e cancels each other)
Please be reminded, this answer may not be correct.