(a) Here the sum of locants in both numbering is 5 (1+4).
(b) Since C = C comes before C \(\equiv\) C in the seniority table, thus the carbon atom bearing C = C bond should be given less number than the carbon bearing C \(\equiv\) C grouping.
(c) Thus the name of the compound is Pent-1-en-4-yne.
