Here's what I would have submitted:

Question 1:

(R1,R2)=1*4+2*5+3*6=32, which is not equal to zero.  Therefore the
columns are not orthogonal.



Question 2:

Applying Gram-Schmidt, we get 
       [1]
b1=R1= [2]
       [3]


b2=R2- <b1,R2>/<b1,b1> * b1 = R2 - 16/7 * b1.
Clearing denominators, b2=7R2-16b1
 [12]                                    [ 4]
=[ 3].  Removing a common factor, set b2=[ 1].
 [-6]                                    [-2]

b3=R3=<b1,R3>/<b1,b1> * b1 - <b2,R3>/<b2,b2> * b2 = 0.

Thus a basis is {b1,b2}
  [1] [ 4]
={[2],[ 1]}
  [3],[-2]



Question 3:

This is <42e1,b1>/<b1,b1>*b1 + <42e1,b2>/<b2,b2>*b2
=3*b1 + 8 * b2 
 [35]
=[14]
 [-7]