Let's first look at the assertion of the mean value theorem:
If [tex]f[/tex] is a (real-valued) function continuous on the closed interval [tex][a,b][/tex] and differentiable on the open interval [tex](a,b)[/tex], then there exists some [tex]c[/tex] within the open interval such that [tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}[/tex].
So the first thing you should always check is that the given function is indeed continuous/differentiable on the given interval. Here, [tex]\sqrt{x-1}[/tex] is defined as long as [tex]x-1\ge0[/tex], or [tex]x\ge1[/tex]. This contains the given interval [tex][1,3][/tex], so we're fine. The derivative is [tex]\dfrac1{2\sqrt{x-1}}[/tex], which also exists for all [tex]x\in(1,3)[/tex], so we're fine on this front as well.
Now, we want to find the precise value(s) of [tex]c[/tex] that are guaranteed to exists by the MVT. All we need to do is plug in what we know to construct an equation to solve for [tex]c[/tex], as is shown in the given solution:
[tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}\iff\dfrac1{2\sqrt{c-1}}=\dfrac{\sqrt{3-1}-\sqrt{1-1}}{3-1}=\dfrac{\sqrt2}2[/tex]
[tex]\implies\sqrt{2(c-1)}=1\implies c=\dfrac32[/tex]
