#### Answer

(a) $h=2$ and $k=0$ (other values for $k$ will work)
(b) $h=0$ and $k=0$ (other values for $h$ and $k$ will work)
(c) $h=2$ and $k=8$

#### Work Step by Step

We know we can find how many solutions a system has by looking at the echelon form of the corresponding augmented matrix.
The augmented matrix of our system is
$$\left[
\begin{array}{ccc}
1 & h & 2\\
4 & 8 & k
\end{array}
\right].$$
To convert this to echelon form, we replace row 2 with _4*(row 1)+(row 2), which gives
$$\left[
\begin{array}{ccc}
1 & h & 2\\
0 & -4h+8 & -8+k
\end{array}
\right].$$
Now we can answer (a), (b), and (c).
(a) For the system to have no solutions (i.e., to be inconsistent), the last column of the augmented matrix must be a pivot column.
Therefore, we must have
$$-4h+8=0$$
and $$ -8+k\neq0.$$
Hence, we must have $$h=2$$
and $$k\neq 8$$
Since any value for $k$ other than $8$ will work, we can choose $k=0.$ Then for $h=2$ and $k=0$, the second row of our matrix is
[0 0 -8], which corresponds to the equation $0=-8,$ which means the system is inconsistent.
(b) For the system to have a unique solution, it must be consistent, which means the last column of the corresponding augmented matrix cannot be a pivot column, and have no free variables.
Now, looking at the echelon form of our augmented matrix, we can see the system will not have any free variables if $-4h+8 \neq0$. Note that if $-4h+8\neq 0$ or equivalently $h\neq2$, then the last column is not a pivot column for any value of $k.$
Since we can choose any value other than 2 for $h$ and any value for $k$, we choose $h=0$ and $k=0.$ Then our matrix becomes
$$\left[
\begin{array}{ccc}
1 & 0 & 2\\
0 & 8 & -8
\end{array}
\right],$$
which corresponds to the system
$$x_1=2\\ 8x_2=-8.$$
Solving for $x_2$, we get $x_2=-1$. Hence our unique solution is $(x_1, x_2)=(2,-1)$.
(c) For the system to have many solutions, it must be consistent and have a free variable.
Looking again at the echelon form of our matrix, we see the system will have a free variable if $-4h+8=0$, or equivalently $h=2$.
Now if $h=2$, then in order for the system to be consistent (i.e., in order for the last column of our matrix not to be a pivot column), we must have $-8+k=0$, or equivalently $k=8$.
So for $h=2$ and $k=8$ our matrix becomes
$$\left[
\begin{array}{ccc}
1 & 2 & 2\\
0 & 0 & 0
\end{array}
\right],$$
Which means our system is consistent and has infinitely many solutions.