Here's what I would have submitted:

Question 1:

chi=det [1-t    2 ]
        [ 2    1-t]
=(1-t)^2 - 4 = t^2 -2t -3.



Question 2:

K^2=[5 4]
    [4 5]
so K^2 -2K -3I=
[5 4] - [2 4] - [3 0]
[4 5] - [4 2] - [0 3]

=[0 0]
 [0 0].
 
In general, chi_A(A)=0.  This is called the Cayley-Hamilton Theorem. 
I may sketch its proof in section at some point.



Question 3:

The eigenvalues of K are the solutions to t^2 -2t -3 =0, or -1 and 3.
The eigenvectors corresponding to 3 are the solutions to K-3I=0, 
or
[-2  2]    [0]
[ 2 -2] x =[0],
multiples of 
[1]
[1].
The eigenvectors corresponding to K=-1 are solutions to K+I=0, or 
[2 2]    [0]
[2 2] x =[0],
multiples of 
[-1]
[ 1].
Putting these together, we get
P=[1 -1] and D=[3  0]
  [1  1]       [0 -1].