The attached figure shows formulation of the problem in this question.
(a) Base camp to the waterwhole to the West
From the figure, this is distance DB.
Angle CBD = 45°
Therefore,
Cos 45 = DB/2 => DB = 2 Cos 45 = 1.41421 km = 1414.21 m
To nearest metre,
DB = 1414 m
(b) Base camp to the waterhole to the North
From the figure, this is CD.
Therefore,
Sin 45 = CD/2 => CD = 2 Sin 45 = 1.41421 km = 1414.21 m
To nearest meter,
CD = 1414 m
(c) Base camp to the initial resting place to the North
From the figure, this is CE.
From geometry;
CE = CD+ DE, but DE = AB = 1.5 km = 1500 m
Then,
CE = 1414+1500 = 2914 m