To solve the given problems, we proceed as follows:

Part (1): Finding the Monthly Growth Rate of Sales

Let's denote the monthly growth rate of sales by $x$. The problem states that the sales increased from 100 helmets in April to 144 helmets in June, with the same monthly growth rate. Since there are two months from April to June, the sales in June can be represented as $100(1+x)^2 = 144$.

To find $x$, we solve the equation:
\begin{align*}
100(1+x)^2 &= 144 \\
(1+x)^2 &= \frac{144}{100} \\
(1+x)^2 &= 1.44 \\
1+x &= \sqrt{1.44} \\
x &= \sqrt{1.44} - 1 \\
x &= 1.2 - 1 \\
x &= 0.2 \text{ or } x = -2.2
\end{align*}

Since a negative growth rate doesn't make sense in this context, we discard $x = -2.2$ and conclude that the monthly growth rate is $20\%$. Thus, $\boxed{x = 20\%}$.

Part (2): Finding the Actual Selling Price for Maximum Profit

Let the actual selling price of the helmet be $m$ yuan. According to the problem, when the selling price is $40$ yuan, the monthly sales volume is 600 helmets. If with each increase of $1$ yuan, the monthly sales volume decreases by 10 helmets, then the sales volume can be represented as $600 - 10(m - 40)$.

The profit is given by the selling price minus the cost price, multiplied by the sales volume. We aim for a profit of $10,000$ yuan, so we set up the equation:
\begin{align*}
(m - 30)[600 - 10(m - 40)] &= 10,000 \\
(m - 30)(600 - 10m + 400) &= 10,000 \\
(m - 30)(1000 - 10m) &= 10,000 \\
1000m - 10m^2 - 30000 + 300m &= 10,000 \\
-10m^2 + 1300m - 40,000 &= 0 \\
m^2 - 130m + 4000 &= 0
\end{align*}

Solving this quadratic equation, we find that $m = 50$ or $m = 80$. To provide customers with the best value, we choose the lower price, which is $m = 50$ yuan per helmet. Therefore, the actual selling price of this brand of helmet should be $\boxed{m = 50}$ yuan per helmet.