Step 1: find eigenvalues. $\chi_A(\lambda) = \det(A-\lambda I) = -\lambda^3+5\lambda^2-8\lambda+4 = -(\lambda-1)(\lambda-2)^2$. We are lucky, all eigenvalues are real.
Step 2: for each eigenvalue $\lambda_\imath$, find rank of $A-\lambda_\imath I$ (or, rather, nullity, $\dim(\ker(A-\lambda_\imath I))$) and kernel itself. For $\lambda=1$:
$$A-\lambda I = \pmatrix{-8 && 8 && 2 \\ -4 && 4 && 1 \\ -23 && 21 && 6}, \ker(A-\lambda I) = L(\pmatrix{3 \\ 1 \\ 8})$$
($L(v_1, v_2, ..., v_n)$ denotes the linear hull of vectors, the set of all their linear combinations.) Algebraic multiplicity of the root is 1, geometric multiplicity is 1, we're done here. For $\lambda=2$:
$$A-\lambda I = \pmatrix{-9 && 8 && 2 \\ -4 && 3 && 1 \\ -23 && 21 && 5}, \ker(A-\lambda I) = L(\pmatrix{2 \\ 1 \\ 5})$$
Algebraic multiplicity of the root is 2, geometric multiplicity is 1. We're unlucky, now we have to solve
$$(A-\lambda I)v=\pmatrix{2 \\ 1 \\ 5} \sim v = \pmatrix{0 \\ 0 \\ 1}$$
Step 3: our matrix in basis $(\pmatrix{3 \\ 1 \\ 8},\pmatrix{2 \\ 1 \\ 5},\pmatrix{0 \\ 0 \\ 1})$ has form $J_A = \pmatrix{1 && 0 && 0 \\ 0 && 2 && 1 \\ 0 && 0 && 2}$. Matrix $P$ corresponding to this basis change is $\pmatrix{3 && 2 && 0 \\ 1 && 1 && 0 \\ 8 && 5 && 1}$, i.e. $P^{-1}AP=J_A$.
Note: If you have a root of algebraic multiplicity 3, but there's only one eigenvector $v_1$, then you seek $v_2:(A−λI)v_2=v_1$ and then $v_3:(A−λI)v_3=v_2$ (note the index!). But when the nullity of $A−λI$ is greater than 1 (and less than algebraic multiplicity), things get a bit tricky. You have to find maximal $k : (A−λI)^k≠0$, then find vector(s) $v_k:(A−λI)^k v_k=0,\,(A−λI)^{k−1}v_k≠0$ (chain generators) and then proceed $v_{k−1}=(A−λI)v_k,v_{k−2}=(A−λI)v_{k−1},...$ up to an eigenvector $v_1$ (see "Jordan chains").
Question 1: Is it true that two matrices are similar if and only if their Jordan forms are equal?
The answer is positive. See Similar Matrices and their Jordan Canonical Forms.
Question 2: Computation of the normal Jordan form
See here.
To complete my answer with what I understood from your question. Yes, The difference between m and k is the number of 1s ABOVE the leading diagonal. However, take care that two matrices with same $m,k$ can be non similar.
Example: two square nilponent matrices of order $6$. One having two Jordan blocks of size $3$, the other one with Jordan blocks of sizes $2$ and $4$.
Best Answer
Let $A$ be any matrix and $A=SJS^{-1}$ its Jordan decomposition. Then, by the definition of the exponential function for matrices, we obtain:
$$ \exp(At)=\exp(SJS^{-1}t) = \sum_{i=0}^{\infty} \frac{(SJS^{-1})^i t^i}{i!} = S\sum_{i=0}^{\infty} \frac{J^it^i}{i!}S^{-1}=S\exp(Jt)S^{-1}$$
since the $S$ cancel in the powers. Now, given $J=\bigoplus_j J_j$ (decomposition of $J$ into Jordan blocks), you have, since the Jordan blocks commute:
$$ \exp(Jt)=\bigoplus_j \exp{J_jt}$$ In particular, if $A$ is diagonalizable, then $\exp(J)$ is also diagonal and the diagonal entries are just $\exp(J_jt)$, with $J_j$ the eigenvalues. Otherwise, you need to calculate the exponential function of the Jordan blocks. This already tells you that there is no "constant c" the way you want to have it. The constant changes with the different Jordan blocks.
Your Jordan blocks will look like $J:=\lambda I + N$ with a diagonal part and an nilpotent part $N$. Since the two parts commute (check this!) we have:
$$\exp(Jt)=\exp(\lambda I t)\exp(Nt)=\exp(\lambda t)\exp{Nt}$$
which might help for calculations.