C is an acute angle, so c = arcsin5 √ 3/14; cosC = 1 1/ 14；

sinB = sin(A+C)= Sina cosc+cosa sinc =(√3/2)× 1 1/ 14+(- 1/2)×(5√3/ 14)= 3√3/ 14

B is an acute angle, so b = arcsin (3 √ 3/14);

And AC/sinB=BC/sinA, so AC = 7× (3 √ 3/14)/(3/2) = 3.

So the three sides of △ABC are: AC=3, AB=5, BC = 7；; Triangle: a = 120, B = arcsine (3 √ 3/14);

c = arcs in 5√3/ 14；

2. because a = 4 >；; B=3, so △ABC has only one solution;

B/sinB = a/ Sina: sinB=(b/a) Sina = 3 √ 2/8; B is an acute angle, so b = arcsin (3 √ 2/8); And cosb = √ 46/8;

sinC = sin(A+B)= Sina cosb+cosa sinb =(√2/2)×(√46/8)+(√2/2)×(3√2/8)=(3+√23)/8；

C is an obtuse angle, so c = π-arcsin [(3+√ 23)/8];

And c/sinc = a/Sina: c = a (sinc/Sina) = 4× (3 √ 2+√ 46)/8 = (3 √ 2+√ 46)/2.

So a=4, b=3, c = (3 √ 2+√ 46)/2; A=45, B = arcsine (3 √ 2/8); π-arcsine [(3+√23)/8].