We are given that $f(x) = \sqrt{x}$.
Hence, the first and second derivatives of $f$ are$f'(x) = \frac1{2\sqrt{x}}$ and f''(x) = -\frac1{4x^{3/2}}. Note that f''(x) < 0 for all $x \in \left(0, \infty\right)$. If the first derivative of a function is always negative, then the function is a decreasing function.
In our case, f''(x) is the first derivative of f'(x) and since f''(x)<0, \forall x \in \left(0, \infty \right), we get that f'(x) is a decreasing function.
For the second part, by mean value theorem, we have that if a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a < b$, and differentiable on the open interval $(a, b)$, then there exists a point $c \in (a, b)$ such that f'(c) = \frac{f(b)-f(a)}{b-a} i.e. f(b) = f(a) + (b-a) \times f'(c). To prove the second part, choose $f(x) = \sqrt{x}$, $a = 100$ and $b=102$. Clearly $f$ satisfies the assumptions in the mean value theorem. Hence, $\exists c \in (a,b)$ such that \sqrt{b} = \sqrt{a} + (b-a) f'(c) = \sqrt{a} + (b-a) \frac1{2 \sqrt{c}}. Plugging in the values for $a$ and $b$ we get that $\sqrt{102} = \sqrt{100} + \frac{2}{2 \sqrt{c}} = 10 + \frac1{\sqrt{c}}$ Since $c \in (100,102)$, we have that $10 < \sqrt{c} <\sqrt{102} < \sqrt{121} = 11$. Hence, $\frac1{11} < \frac1{\sqrt{c}} < \frac1{10}$. Hence, $10 + \frac1{11} < 10 + \frac1{\sqrt{c}} < 10 + \frac1{10}$ i.e. $10 + \frac1{11} < \sqrt{102} < 10 + \frac1{10}$.