(1) Let A be AD⊥BC, and the extension line of BC is at point D, as shown in figure 1:

In Rt△ADC, AC=4,

∫∠C = 150，

∴∠ACD=30，

∴AD= 1/2AC=2，

CD=AC？ cos30 =4×√3/2=2√3，

In Rt△ABD, tanB=AD/BD=2/BD= 1/8,

∴BD= 16，

∴bc=bd-cd= 16-2√3；

(2) Take a little m on the side of BC, so that CM=AC, and then connect AM, as shown in Figure 2:

∫∠ACB = 150，

∴∠AMC=∠MAC= 15，

tan 15 = tan∠AMD = AD/MD = 2/(4+2√3)= 1/(2+√3)≈ 1/(2+ 1.7)≈0.27≈0.3