True or false: Every real homogeneous linear system of equation which has more than one solution has infinite solutions

Indeed, this is even true for non-homogeneous linear systems. Consider the system $Ax=b$, and assume $x_0$ and $x_1$ are solutions. Then for any $x_\lambda = (1-\lambda)x_0+\lambda x_1$ you get $$Ax_\lambda = A((1-\lambda)x_0 + \lambda x_1) = (1-\lambda)A x_0 + \lambda A x_1 = (1-\lambda) b + \lambda b = b$$ Therefore $x_\lambda$ is also a solution, thus you get infinitely many (indeed even uncountably many) solutions.

The homogeneous system is just the special case for $b=0$. Since $x=0$ is always a solution of a homogeneous linear system, for those you can even write the condition as:

If any real homogeneous linear system of equations has a non-zero solution, it has infinitely many.

Hint: if the homogeneous system has two distinct solutions, then one of them, call it $v$, is nonzero. What can you say about $\alpha v$ for a scalar $\alpha$?

Yes indeed this is true. The way that this is best seen is by noting that if a homogeneous system of equations has more than one solution, then the matrix corresponding to the system of equations, $A$, has a non-trivial kernel.

This means that $\exists \vec{v},\vec{u} \in \ker(A): A\vec{v} = 0$ and $A\vec{u} = 0$. Now it is worth noting that we can take any linear combination of these vectors $\lambda \vec{v} + \mu \vec{u}$ for $\lambda, \mu \in \mathbb{R}$ and we have $A(\lambda \vec{v} + \mu \vec{u}) = \lambda A\vec{v} + \mu A\vec{u} = 0$, so $\lambda \vec{v} + \mu \vec{u}$ also solves this system of equations.

Therefore we have an infinite number of solutions if we have more than one unique solution.