Why can't $\int_1^3 \frac{4}{(2x-3)^4} dx $ be evaluated by calculators or WolframAlpha?

OP: the interval change still does explain why the integral cannot be solved on calculators

This is because the "dysfunction"/discrepancy that you are raising has nothing to do with the integration-by-substitution process.

If $f$ is Riemann integrable on $[a,b]$ and has antiderivative $F,$ the second fundamental theorem of calculus guarantees that $$\int_a^b f=F(b)-F(a).$$ In your example, the boldfaced condition is not satisfied (the integral fails to converge/exist due to the singularity at $x=1.5$), and $$\int_1^3 \frac{4}{(2x-3)^4} \mathrm dx\ne \left[\frac2{3 (3 - 2 x)^3}\right]_1^3,$$ even though $$\int\frac{4}{(2x-3)^4} \mathrm dx=\frac2{3 (3 - 2 x)^3}+C.$$

This, together with Martin's example in the second comment, shows that an integrand may have an antiderivative whilst not being integrable on the desired interval.

When we reason$$\int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a),$$we assume every $g$ with $g^\prime=f$ satisfies $g(b)-g(a)=F(b)-F(a)$, so the choice of $f$'s antiderivative is irrelevant.

But in general, antiderivatives differ additively by locally constant functions on the integrand's domain, which due the discontinuity at $x=\frac32$ is in this case in two components as $[1,\,3]\setminus\left\{\frac32\right\}=\left[1,\,\frac32\right)\cup\left(\frac32,\,4\right]$. We can make this even more technical, but that's all we need for now.

A locally constant function on this domain can have different values either side of $\frac32$, say $C_-$ on the left and $C_+$ on the right. But if we add such a function to an antiderivative of $4(2x-3)^{-4}$, the difference between the antiderivatives' values at $a,\,b$ with $a<\frac32<b$ changes by $C_+-C_-$, which is in general nonzero, and so the integral is improper.