Let $z=x+iy$.

We interpret the two loci in the Argand plane.

1. First locus: $|z-(4+3i)|=2$

This is a circle with center $(4,3)$ and radius $2$.

So its equation is

$$(x-4)^2+(y-3)^2=4.$$

2. Second locus: $|z|+|z-4|=6$

Here,

• $|z|$ is distance from $(0,0)$,
• $|z-4|$ is distance from $(4,0)$.

Hence this is an ellipse with foci $(0,0)$ and $(4,0)$, and sum of distances equal to $6$.

Since for an ellipse, $2a=6 \implies a=3,$ and distance between foci is $4$, so $2c=4 \implies c=2.$ Thus, $b^2=a^2-c^2=9-4=5.$

The center is $(2,0)$, so the ellipse is

$$\frac{(x-2)^2}{9}+\frac{y^2}{5}=1.$$

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3. Solve simultaneously

We need intersections of

$$(x-4)^2+(y-3)^2=4 \tag{1}$$

and

$$\frac{(x-2)^2}{9}+\frac{y^2}{5}=1. \tag{2}$$

Instead of solving fully algebraically, use geometry first.

The circle center is $(4,3)$ with radius $2$. The ellipse is symmetric about the $x$-axis.

Check whether the circle can meet the lower half of the ellipse:

• On the circle, lowest point is at $y=1$.
• So any intersection must have $y\ge 1$. Thus only upper branch of ellipse matters.

Now parametrize the circle:

$$x=4+2\cos\theta, \qquad y=3+2\sin\theta.$$

Substitute into ellipse:

$$\frac{(2+2\cos\theta)^2}{9}+\frac{(3+2\sin\theta)^2}{5}=1.$$

This is cumbersome, so let us instead compare positions.

4. Use upper branch of ellipse

From ellipse:

$$y=\sqrt{5\left(1-\frac{(x-2)^2}{9}\right)}.$$

From circle (lower branch gives $y<3$, upper branch gives larger values, but circle itself lies between $y=1$ and $y=5$):

$$y=3\pm\sqrt{4-(x-4)^2}.$$

Since ellipse upper branch has $y\ge 0$, intersections with the circle occur where

$$\sqrt{5\left(1-\frac{(x-2)^2}{9}\right)}=3-\sqrt{4-(x-4)^2}$$

or

$$\sqrt{5\left(1-\frac{(x-2)^2}{9}\right)}=3+\sqrt{4-(x-4)^2}.$$

Geometrically, the ellipse's top is at $(2,\sqrt5)\approx(2,2.236)$, while the circle center is above and right of this. The circle spans $x\in[2,6]$.

Let us test key points:

• At $x=2$, circle gives

$$(2-4)^2+(y-3)^2=4 \implies 4+(y-3)^2=4 \implies y=3.$$

Ellipse gives

$$\frac{0}{9}+\frac{y^2}{5}=1 \implies y=\pm\sqrt5.$$

No intersection.

• At $x=4$, circle gives

$$y=3\pm 2 \implies y=5,1.$$

Ellipse gives

$$\frac{(4-2)^2}{9}+\frac{y^2}{5}=1 \implies \frac49+\frac{y^2}{5}=1 \implies y^2=\frac{25}{9} \implies y=\pm\frac53.$$

So at $x=4$, ellipse point lies between the two circle points, suggesting crossing.

Now compare functions on interval $x\in[2,4]$ and $[4,5]$.

Define

$$f(x)=(x-4)^2+\left(\sqrt{5\left(1-\frac{(x-2)^2}{9}\right)}-3\right)^2-4.$$

Intersections correspond to $f(x)=0$.

Compute signs:

• At $x=2$:

$$f(2)=4+(\sqrt5-3)^2-4=(\sqrt5-3)^2>0.$$

• At $x=4$:

$$f(4)=0+\left(\frac53-3\right)^2-4 =\left(-\frac43\right)^2-4 =\frac{16}{9}-4=-\frac{20}{9}<0.$$

So there is at least one root in $(2,4)$.

• At $x=5$, ellipse gives

$$\frac{9}{9}+\frac{y^2}{5}=1 \implies y=0.$$

Then

$$f(5)=(1)^2+(0-3)^2-4=1+9-4=6>0.$$

Since $f(4)<0$ and $f(5)>0$, there is at least one root in $(4,5)$.

Thus there are at least $2$ intersections.

Because the upper branch of the ellipse is single-valued and the circle arc over $x\in[2,5]$ can cross it at most twice here, total intersections are exactly $2$.

Therefore, the number of points of intersection is

$$\boxed{2}.$$

So the correct option is C.