From the comments, it seems that your first problem is supposed to read something like:
Suppose $A$ and $B$ are matrices, and $\vec{b}$ is such that the systems of linear equations $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ have infinitely many solutions in common. Is it true that $(A+B)\vec{t}=\vec{b}$ has infinitely many solutions?
If that's the case, then your argument works: for each common solution $\vec{x}$ to the original two systems, $\frac{1}{2}\vec x$ is a solution to $(A+B)\vec{x}=\vec{b}$; since \frac{1}{2}\vec{x} = \frac{1}{2}\vec{x'} if and only if \vec{x}=\vec{x'}, it follows that the latter system has infinitely many solutions as well.
But you need the solutions to be common solutions, as Yuval's example shows.
The second question is false in two interpretations:
Suppose $A$ and $B$ are matrices, and $\vec{b}$ is such that the systems of linear equations $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ have no solutions in common. Is it true that $(A+B)\vec{x}=\vec{b}$ has no solutions?
and
Suppose that $A$ and $B$ are matrices, and $\vec{b}$ is such that the systems of linear equations $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ each has no solutions. Is it true that $(A+B)\vec{x}=\vec{b}$ has no solutions?
This is false; note that any pair of matrices $A$ and $B$ and vector $\vec{b}$ that satisfy the second statement will also satisfy the first, so it suffices to find a counterexample to the second statement. The following works: $A = \left(\begin{array}{cc}1&0\\0&0\end{array}\right),\quad B=\left(\begin{array}{cc}0&0\\0&1\end{array}\right),\quad \vec{b}=\left(\begin{array}{c}1\\1\end{array}\right).$
Suppose now we tweak it a bit, perhaps; how about the following?
Suppose $A$ and $B$ are matrices, and $\vec{b}$ is such that $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ each has solutions, but there are no solutions in common to both systems. Is it true that $(A+B)\vec{x}=\vec{b}$ has no solutions?
This is still not true. Take $A=\left(\begin{array}{crc} 1&-1&0\\ 0&0&1 \end{array}\right),\quad B=\left(\begin{array}{ccr} 1&0&0\\ 0&1&-1 \end{array}\right),\quad \vec{b}=\left(\begin{array}{c}1\\1\end{array}\right).$ Then $A\vec{x}=\vec{b}$ has solutions: $\vec{x}=(1,0,1)^T$ is a solution. $B\vec{x}=\vec{b}$ also has solutions: $\vec{x}=(1,1,0)^T$ is a solution.
But $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ have no solutions in common: if $\vec{x}=(x,y,z)^T$ were a solution, then you would need $x-y=1$, $z=1$, $x=1$, and $y-z=1$. But from $x=z=1$ and $x-y=0$, we get $y=0$; and from $z=1$ and $y-z=1$ we get $y=2$.
However, $(A+B)\vec{x} = \left(\begin{array}{rrr}2 & -1 & 0\\ 0 & 1 & 0 \end{array}\right)\vec{x} = \left(\begin{array}{c}1\\1\end{array}\right)$ does have solutions: $(1,1,z)^T$ is a solution for all $z$.