### Solution:

#### Part (1): Domain and Range of $f(x)$

Given $f(x)=\sqrt{1+x}+\sqrt{1-x}$, we need to ensure the expressions under the square roots are non-negative for the function to be real-valued. Therefore, we have two inequalities:

1. $1+x\geqslant 0 \Rightarrow x\geqslant -1$
2. $1-x\geqslant 0 \Rightarrow x\leqslant 1$

Combining these, we find the domain of $f(x)$ is $-1\leqslant x\leqslant 1$.

To find the range, we consider $y=f(x)=\sqrt{1+x}+\sqrt{1-x}$. Squaring both sides gives us:

$$y^2 = (\sqrt{1+x}+\sqrt{1-x})^2 = 2 + 2\sqrt{1-x^2}$$

Since $0\leqslant \sqrt{1-x^2}\leqslant 1$, we have:

$$2\leqslant y^2 \leqslant 4$$

Taking square roots and considering $y\geqslant 0$, we get:

$$\sqrt{2}\leqslant y\leqslant 2$$

Therefore, the domain of $f(x)$ is $\boxed{[-1,1]}$ and the range is $\boxed{[\sqrt{2}, 2]}$.

#### Part (2): Range of Real Number $m$

Given $h(x)=\frac{f(x)}{2\sqrt{1-x^2}+6}$, we can express $h(x)$ in terms of $t=\sqrt{1+x}+\sqrt{1-x}$, where $t\in[\sqrt{2},2]$. Then, $2\sqrt{1-x^2}=t^2-2$, and we consider the function $g(t)=\frac{t}{t^2+4}=\frac{1}{t+\frac{4}{t}}$.

By AM-GM inequality, $t+\frac{4}{t}\geqslant 2\sqrt{t\times\frac{4}{t}}=4$, with equality when $t=2$. Thus, the maximum value of $g(t)$, and hence $h(x)$, is $\frac{1}{4}$.

The inequality $h(x)\leqslant \frac{3}{4}m^2-\frac{1}{2}am$ transforms into $\frac{3}{4}m^2-\frac{1}{2}am\geqslant\frac{1}{4}$ for any $a\in[-1,1]$. This simplifies to:

$$3m^2\pm am-1\geqslant 0$$

Considering the cases for $a=\pm 1$, we get two inequalities:

1. $3m^2+m-1\geqslant 0$
2. $3m^2-m-1\geqslant 0$

Solving these inequalities, we find:

1. $m\leqslant -1$ or $m\geqslant \frac{1}{3}$
2. $m\leqslant -\frac{1}{3}$ or $m\geqslant 1$

Combining these results, we conclude that $m\leqslant -1$ or $m\geqslant 1$. Therefore, the range of the real number $m$ is $\boxed{(-\infty, -1]\cup[1, +\infty)}$.