For no solution, the determinant of the coefficient matrix must be zero but the system must be inconsistent.

$\Delta = (k+1)(k+3) - 8k = k^2+4k+3-8k = k^2-4k+3 = (k-1)(k-3) = 0$

So $k=1$ or $k=3$. Check each:

$k=1$: Equations become $2x+8y=4$ and $x+4y=2$ — same line (infinite solutions). Not "no solution".

$k=3$: Equations become $4x+8y=12$ and $3x+6y=8$. Simplify: $x+2y=3$ and $x+2y=8/3$ — contradiction! No solution.

So exactly $\mathbf{1}$ value of $k$ gives no solution.