1)△ABC, where c = √ 3asinc-cccosa.

According to sine theorem, a/sinA=b/sinB=c/sinC=2R.

Substitute the conditional expression and omit 2R:

sinC =√3 Sina sinC-sinC cosa & gt； 0

So: √3sinA-cosA= 1.

So: cosA=√3sinA- 1

Substitute into cos? A+ sin? A= 1:

1-2√3sinA+3sin？ A+ sin? A= 1

So: 2(2sinA-√3)sinA=0.

Solution: sinA=√3/2(sinA=0 does not conform to giving up)

Solution: A = 60 or A = 120.

Obviously, A = 120 does not satisfy cosA=√3sinA- 1.

To sum up, a = 60

2)a=2，S△ABC=(bc/2)sinA=√3

So: BCS in 60 years = 2 √ 3.

Solution: bc=4

According to cosine theorem: cosA=(b? +c？ -a？ )/(2bc)= 1/2

Yes: b? +c？ -4=4，b？ +c？ =8

Solution: b=c=2