Is there any short trick that will give the minimal polynomial? 
The minimal polynomial of
\begin{bmatrix} 2& 8&0&0&0&0&0 \\ 0&2&0&0&0&0&0 \\ 0&0&4&0&0&0&0 \\ 0&0&1&3&0&0&0 \\ 0&0&0&0&0&3&0 \\ 0&0&0&0&0&0&0 \\ 0&0&0&0&0&0&5\end{bmatrix}
is
$(1)  \ t(t-2)(t-5) \\ (2) \ t(t-2)^2(t-5) \\ (3) \ t^2(t-2)(t-5) \\ (4) \ t^2(t-2)^2(t-5)$.

If I expand the determinant $(A-\lambda I)$, then I will get with more calculation.
But is there any short trick that will give the minimal polynomial ?
 A: None of the options are correct : this can be seen from the fact that $4$ is an eigenvalue with eigenvector $[0,0,1,1,0,0,0]$. Therefore, since any eigenvalue must be a root of the minimal polynomial, we get that $(t-4)$ must be a factor of the minimal polynomial. This is not the case with any of the given polynomials.

The idea is that the given matrix(call it $A$) is a block matrix (with blocks $A_i$, say) : therefore, if $p$ is any polynomial, then $p(A)$ is also a block matrix with blocks $p(A_i)$. In particular, the matrix $p(A)$ is zero exactly when each of the individual blocks are zero.
We know when the blocks are zero : for this, $p$ must be a multiple of the minimal polynomial of each of the blocks. Naturally, with this in mind, the minimal polynomial of $A$ is seen to be the lcm , of the minimal polynomials of the blocks.
Now, to calculate the lcm of the blocks, let us first see what the blocks are: they are $\begin{bmatrix} 2 \quad 8 \\ 0 \quad 2 \end{bmatrix} $,$\begin{bmatrix} 4 \quad 0 \\ 1 \quad 3 \end{bmatrix}$ , $\begin{bmatrix} 0\quad 3 \\ 0 \quad 0 \end{bmatrix}$ and $[5]$.   
Each of these has minimal polynomial : $(t-2)^2$, $(t-4)(t-3)$, $t^2$ and $t-5$ respectively. These four cases are seen from the fact that all the matrices are either upper or lower triangular, so it is easy to see their eigenvalues and thence the minimal polynomial.The LCM is clearly seen to be $t^2(t-2)^2(t-3)(t-4)(t-5)$. 
