Is this Integration From First Principles notation standard outside of A Level Maths?

In my experience as a UK mathematician, this isn't standard notation. The aim is to convey $x$'s step size is $\delta x$ rather than $1$. I'd recommend writing the limit as $\lim_{\delta x\to0}\sum_{k=0}^{5/\delta x}\sqrt{4+k\delta x}\delta x$ - no, wait, make it $\lim_{N\to\infty}\sum_{k=0}^N\sqrt{4+5k/N}(5/N)$. This limit is the stated integral; in fact, if we set the maximum $k$ to $N-1$ instead (which won't change the limit), it's a Riemann sum.

1. $$\lim_{\delta x \to0}\sum_{r = 1}^n f(x_r^*)\,\delta x$$ and $$\lim_{n \to\infty}\sum_{r = 1}^n f(x_r^*)\,\delta x$$ are equivalent expressions (containing Riemann sums).

2. Loosely generalising the index of summation to $x$, $$\lim_{\delta x \to0}\sum_{x = x_1^*}^{x_n^*} f(x)\,\delta x,$$ I reckon that the expression remains conceptually sound.

3. If the limits of integration are $a$ and $b$, $$\lim_{\delta x \to0}\sum_{x = a}^{b-\delta x} f(x)\,\delta x$$ (containing a right Riemann sum) is equivalent to the above expression.

4. $$\lim_{\delta x \to0}\sum_{x = a}^{b} f(x)\,\delta x$$ converges to the same value, I think. This is the notation used in the syllabus/specification in question.

   However, the sum now ceases to be a Riemann sum, as we are now considering $(n+1)$ instead of just $n$ subintervals. I don't find this hand-waving beneficial to thoughtful learners, because it conflicts with the intuitive partition of $n$ rectangles of (in our case fixed) width $\delta x$.

Given that the specification (p. 25) requires only that the candidate recognise the limit of a sum as an integral, and the exam (Q5) accordingly just expects the candidate to rewrite $\displaystyle \lim_{\delta x \to 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x\,\,\text{ as }\,\displaystyle \int_4^9\sqrt{x} \,\mathrm{d}x$ $\,\,\big(\text{for 1 mark}\big),$ I think the nonstandard notation is acceptable (albeit not particularly instructive).

ADDENDUM

$\delta x$ is the width of the $n$ subintervals of our regular partition, i.e., $\displaystyle\delta x=\frac{b-a}n$, so $\delta x$ and $n$ vary in tandem as the Riemann sum approaches the definite integral.

Your error in $\boxed{\lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x = \lim_{\delta x \rightarrow 0}\left( \sqrt{4}(\delta x) + \sqrt{5} (\delta x) + \cdots + \sqrt{9} (\delta x)\right)=0}$ was in choosing a step size of 1 instead of letting it vary as $\delta x$. (After all, with the index of summation generalised from $r$ to $x$, the step size ought to be correspondingly changed from $1$ to $\delta x$.) This is why the first equality in the box is false.