解：（1）接收到超声波时下潜的时间$t = t_1 + t_2 = 60\ s + 4\ s = 64\ s，$则下潜的深度即其距离水面的深度$s_1 = v_{奋斗者}t = 1\ m/s×64\ s = 64\ m。$
（2）超声波$4\ s$内传播的路程$s_2 = v_{超声波}t_2 = 1500\ m/s×4\ s = 6000\ m，$从发射到接收到超声波奋斗者号下潜的距离$s_3 = v_{奋斗者}t_2 = 1\ m/s×4\ s = 4\ m，$刚发出超声波时奋斗者号距离海底的高度$h = \frac{1}{2}(s_2 + s_3) = \frac{1}{2}×(6000\ m + 4\ m) = 3002\ m。$
（3）水面到海底的深度$h_{总} = h + v_{奋斗者}t_1 = 3002\ m + 1\ m/s×60\ s = 3062\ m，$则从水面下潜到海底消耗的时间$t_{总} = \frac{h_{总}}{v_{奋斗者}} = \frac{3062\ m}{1\ m/s} = 3062\ s。$
