Note that for all real $k\ge0$ and $a>0$ we have
$$\frac1{k+a}=\int_0^1 x^k x^{a-1}\,dx.
$$
Differentiating in $k$, we further get
$$\frac1{(k+a)^2}=-\int_0^1 x^k \ln x\, x^{a-1}\,dx.
$$
So, doing the partial fraction decomposition, we have
$$\frac2{3(k+2)}+\frac8{(k+1)(k+2)^2(k+3)}
=\frac2{3(k+2)}-\frac4{k+3}+\frac4{k+1}-\frac8{(k+2)^2}
=\int_0^1 x^k f(x)\,dx=EX^k
$$
if the pdf of $X$ is given by
$$f(x):=4+\frac{2x}3-4 x^2+8x\ln x \tag{1}
$$
for $x\in(0,1)$.

Note also that any compactly supported distribution on $\mathbb R$ is uniquely determined by its moments, because then the characteristic function of the distribution is uniquely determined by the moments; here one can use the moment generating function instead of the characteristic one.

So, here the distribution of $X$ with its pdf $f$ given by (1) is uniquely determined by the moments.

Note further that the function $f$ defined by (1) is indeed a pdf. First here, $\int_0^1 f(x)\,dx=1$. Also, $f''(x)=8/x-8>0$ for $x\in(0,1)$ and $f'(3/5)=-0.219\ldots<0$. So, $f$ is convex on $(0,1)$ and hence
$$f(x)\ge f(3/5)+f'(3/5)(x-3/5)\ge f(3/5)+f'(3/5)(1-3/5)
=0.42006\ldots>0
$$
for all $x\in(0,1]$.
Here is the graph of $f$:

This method should work for any compactly supported distribution on $\mathbb R$ whose $k$th moments are given by a rational (in $k$) expression.