Proof:

Part (1): To show $2x^{2}+5x+3 > x^{2}+3x+1$ for $\forall x\in \mathbb{R}$,

First, subtract the right-hand side from the left-hand side:

$$\begin{align*} (2x^{2}+5x+3)-(x^{2}+3x+1) & = 2x^2 + 5x + 3 - x^2 - 3x - 1 \\ & = x^{2} + 2x + 2 \\ & = (x + 1)^{2} + 1. \end{align*}$$

Since for $\forall x\in \mathbb{R}$, we have $(x+1)^{2} \geq 0$, thus $(x + 1)^{2} + 1 > 0$.

Therefore, it follows that $2x^{2}+5x+3 > x^{2}+3x+1$.

Part (2): To show $\sqrt{a} > \sqrt{b}$ for $a > b > 0$,

Consider the difference $\sqrt{a}-\sqrt{b}$:

$$\begin{align*} \sqrt{a}-\sqrt{b} & = \frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}} \\ & = \frac{a - b}{\sqrt{a} + \sqrt{b}}. \end{align*}$$

Given that $a > b > 0$, it implies $a - b > 0$. Also, since $a > 0$ and $b > 0$, $\sqrt{a} + \sqrt{b} > 0$.

Therefore, we have $\frac{a-b}{\sqrt{a}+\sqrt{b}} > 0$, which implies $\sqrt{a} > \sqrt{b}$.

Thus, the proofs for both parts are as follows:

For part (1): $2x^{2}+5x+3 > x^{2}+3x+1$ is true for all $x \in \mathbb{R}$.

For part (2): $\sqrt{a} > \sqrt{b}$ for $a > b > 0$.

Final conclusions are:

- For part (1): $2x^{2}+5x+3 > x^{2}+3x+1$ $\boxed{\text{True}}$
- For part (2): $\sqrt{a} > \sqrt{b}$ $\boxed{\text{True}}$