Please help!!! It is F5 Maths, 3D problem. The numbers next to each subparts of the question are the answers to those subparts.?

Let E to be the origin ,
and EF be x-axis , EH be y-axis , EA be z-axis .
Then we have the following coordinates:
A(0,0,8) , F(8,0,0) , G(8,8,0) , H(0,8,0) , M(0,4,4)

(a)
FM
= ∣(-8,4,4)∣
= √ [ (-8)^2+4^2+4^2 ]
= 4 √ [ 2^2+1^2+1^2 ]
= 4√6 ...Ans

(b)
N
= A + s( vector AG ) , for some s
= (0,0,8) + s(8,8,-8)
= (0,0,8) + t(1,1,-1) , let t = 8s
= (t,t,8-t)
So, NM = (-t,4-t,-4+t)

NM．AH
= (-t,4-t,-4+t)．(0,8,-8)
= 8(4-t) - 8(-4+t)
= 64 -16t
= 0
Hence, t = 4 , N = (4,4,4)

FN
=∣(-4,4,4)∣
= √ [ (-4)^2+4^2+4^2 ]
= 4 √ [ 1^2+1^2+1^2 ]
= 4√3 ...Ans

(c)
FA = (-8,0,8) , FH = (-8,8,0)
FA×FH = (-64,-64,-64) , where × is "cross product"
And (-64,-64,-64) is the vector perpendicular to the plane AFH
GA = (-8,-8,8) , GH = (-8,0,0)
GA×GH = (0,-64,-64) , the vector perpendicular to the plane AGH
Suppose the angle between the plane AFH and AGH is θ
(-64,-64,-64)．(0,-64,-64) = ∣(-64,-64,-64)∣ ∣(0,-64,-64)∣ cosθ
2 * 64^2 = 64√3 * 64√2 * cosθ
cosθ = 2 / √6 = √6 / 3

θ
= arccos (√6 / 3 )
≒ 0.615479708670387 rad
= ( 0.615479708670387 * 180 / π )°
≒ 35.2643896827547 ° ...Ans

a)
FM² + AM² = AF²
FM² + 8²/2 = 8² + 8²
FM² = 96
FM = 4√6

b)
NM ⊥ AH and GH ⊥ AH so NM // GH then N is the mid point of AG.
(2FN)² = DH² + FH²
4FN² = 8² + (8² + 8²)
FN² = √48
FN = 4√3

c)
The required angle = ∠NMF.
cos∠NMF = (NM² + FM² - FN²) / (2NM × FM)
cos∠NMF = (4² + 96 - 48) / (2 × 4 × 4√6) = 2/√6
∠NMF = 35.264... = 35.3° (3 sig. figures)