Maths problem, how to do (a), (b)1 thanks.?

19(a)
7+2√12 = 4+3+2√4√3 = (√4+√3)² = (2+√3)² so a=2.

(b)(i)
Putting x = -1/3, 
3k(1/9-1/3-2) - (-1-19) = 0
k(1-3-18) + 60 = 0
k = 3

(b)(ii)
Putting x = 1,
-(3-19) = (4p)²
16 = 16p²
p² = 1
Hence, 9(x²+x-2) - (3x-19) = 4√3+7
(3px+p)² = 4√3+7
p²(3x+1)² = (2+√3)² 
(3x+1)² = (2+√3)² 
3x+1 = 2+√3 or 3x+1 = -2-√3
x = (1+√3)/3 or x = (-3-√3)/3