The number of values of $k$, for which the system of equations$(k+1)x+8y=4k$$kx+(k+3)y=3k-1$ has no solution, is:

For a system of two linear equations $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$ to have no solution, the condition is:
$$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$$
$\frac{k+1}{k}=\frac{8}{k+3}$$(k+1)(k+3)=8k\Rightarrow k^2+4k+3=8k\Rightarrow k^2-4k+3=0$$(k-1)(k-3)=0\Rightarrow k=1,3$.Check $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$:If $k=1$: $\frac{8}{4}=2$ and $\frac{4k}{3k-1}=\frac{4}{2}=2$. Since all ratios are equal, there are infinite solutions.If $k=3$: $\frac{8}{6}=\frac{4}{3}$ and $\frac{4k}{3k-1}=\frac{12}{8}=\frac{3}{2}$. Here $\frac{4}{3}\ne \frac{3}{2}$.Thus, only $k=3$ results in no solution. The number of values is $1$.