quick high-level overview
Raquel, One of the problems we each face at some point when first learning about parts of the world we can't directly touch and feel with our own senses is about "breaking" wrong ideas we previously held, replacing them with better ideas.
One of the good ideas is to think of a voltage source as a short and to think of a current source as an open.
Let's see why:
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simulate this circuit– Schematic created using CircuitLab
On the left, before pressing the switch, there will be exactly \$10\:\text{mA}\$ in \$R\$. Same on the right, too.
But the moment we press the switch on the left the added \$5\:\text{mA}\$ goes straight through \$V_s\$ and bypasses \$R\$, entirely. \$V_s\$ acts like a short to this added current, all of which passes through \$V_s\$ to ground. \$R\$'s current stays perfectly the same as before. No change at all. So \$V_s\$ acts like a short to the newly added current.
It's even clearer to see this fact if you imagine that \$V_s\$ is a \$0\:\text{V}\$ voltage source. In that case, it's more obviously a short. All a voltage source does is to hold up some voltage difference between two nodes. This might be \$0\:\text{V}\$ or it might be some non-zero value. But regardless of what that value happens to be, it is effectively a dead short!
Hopefully, this helps you see that a voltage source is a short circuit, in some fashion.
The moment we press the switch on the right the added \$5\:\text{mA}\$ goes straight through \$R\$ and bypasses \$I_s\$, entirely. \$I_s\$ acts like an open to this added current, none of which passes through \$I_s\$ to ground instead forcing all of it to be added to \$R\$'s current. \$I_s\$'s current stays perfectly the same as before. No change at all. So \$I_s\$ acts like an open to the newly added current.
It's even clearer to see this fact if you imagine that \$I_s\$ is a \$0\:\text{A}\$ current source. In that case, it's more obviously an open. All a current source does is inject a current into a node. This might be \$0\:\text{A}\$ or it might be some non-zero value. But regardless of what that value happens to be, it is effectively doing so completely immune to any attempts to alter that value. Being immune is the same thing as having no conductance -- or infinite impedance.
Hopefully, this helps you see that a current source is an open circuit, in some fashion.
This is a way of understanding why a voltage source can be treated as a short and why a current source can be treated as an open.
your cases
It's easier to see your schematic as the following:
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Each side can be reduced to the following Thevenin equivalents:
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Here, it is pretty easy to see that the difference voltage will be about \$502.283\:\text{mV}\$ and the resistance will be the sum, or about \$2.4338\:\text{k}\Omega\$.
So that is what anything attached between \$a\$ and \$b\$ will see.
your instructions
One could either use a voltage source between \$a\$ and \$b\$ or a current source. Either method works fine. If you apply a voltage source, then measure the current, that gets the same result as if you apply a current source and measure the voltage difference. Regardless, the voltage difference divided by the current difference gets you the Thevenin impedance.
So there isn't just one way (using a voltage source) but actually two different ways. Both get you to the same place, though.
In the open-circuit case (removing \$R_{_\text{L}}\$), measuring the voltage difference between \$a\$ and \$b\$ will get you the Thevenin voltage. So that instruction gets the right result there.