A closed form for $\int_0^\pi \lvert \sin(m t) \cos(n t) \rvert \, \mathrm{d} t$
Case $m = 1$
By using Maple, we have \begin{align} f(\tfrac{1}{n}) &= \frac{1}{\pi}\int_0^\pi |\sin t \cos (nt)| \mathrm{d} t\\ &= \frac{2(n - \sin \tfrac{\pi}{2n})}{(n^2-1)\pi \sin \tfrac{\pi}{2n}}. \end{align} We have $$\lim_{n\to \infty} f(\tfrac{1}{n}) = \frac{4}{\pi^2}.$$
Case $m=2$, and $n$ odd
By using Maple, we have \begin{align} f(\tfrac{2}{n}) &= \frac{1}{\pi}\int_0^\pi |\sin 2t \cos (nt)| \mathrm{d} t\\ &= \frac{2(n \cos \frac{\pi}{2n}-2\sin \frac{\pi}{2n} )}{(n^2-4)\pi \sin \frac{\pi}{2n}}. \end{align} We have $$\lim_{n\to \infty} f(\tfrac{2}{n}) = \frac{4}{\pi^2}.$$
General Case $m \ne n$
Based on the results for $m=1, 2, 3, 4, \cdots$ by Maple, after simplification and observation, I $\color{blue}{\textrm{GUESS}}$ that, for $m, n \ge 1$ and $m\ne n$, \begin{align} &\int_0^\pi |\sin (mt) \cos (nt)| \mathrm{d} t \\ =\ & \frac{2n}{n^2-m^2} \left(\sum_{s=0}^{\lfloor \frac{n}{2} + \frac{3}{4}\rfloor - 1} \left|\sin \frac{m\pi (1 + 4s)}{2n}\right| + \sum_{s=0}^{\lfloor \frac{n}{2} + \frac{1}{4} \rfloor - 1} \left|\sin \frac{m\pi (3 + 4s)}{2n}\right|\right)\\ &\quad - \frac{2m}{n^2-m^2}\sum_{s=0}^{m-1} \left|\cos \frac{s\pi n}{m}\right| \end{align} where $\lfloor x \rfloor$ is the floor function.
Remarks: 1. I have done numerical experiments $1\le m, n \le 10$ and more.
Rigorous and step-by-step proofs are expected.
Further simplification may be possible.
Notation
$$f_{m,n}(x)=\sin(mx)\cos(nx)\\ \pi I_{m,n}=\int_0^{\pi}dx |f_{m,n}(x)|\\ \mathcal{I}_{m,n,k}=\int_{\frac{\pi (2k-1)}{2n}}^{\frac{\pi (2k+1)}{2n}}f_{m,n}(x)dx\\ \mathcal{I}_{m,n,0}=\int_0^{\pi/(2n)}f_{m,n}(x)dx\\ \mathcal{I}_{m,n,n}=\int^{\pi}_{\pi-\pi/(2n)}f_{m,n}(x)dx $$
Lemma 1:
$$ \int dt \cos(nt)\sin(mt)=\frac{1}{n^2-m^2}(m\sin(nt)\sin(mt)+n\cos(nt)\cos(mt)) $$
Proof: direct differentation
Partial results I: $m=1$, $n \in \mathbb{N}$.
Define $$ I_{1,n}=I_{n}\\ f_{1,n}(x)=f_{n}(x)\\ \mathcal{I_{k,n}}=\mathcal{I_{1,k,n}} $$ Now,we can split the integral as follows: $$ \pi I_n= \mathcal{I}_{n,0}+\sum_{k=1}^{n-1} (-1)^k \mathcal{I}_{n,k}+(-1)^n\mathcal{I}_{n,n}\quad(\star) $$
By Lemma 1, we have by taking into account that $k,n \in \mathbb{N}$ that $$ \mathcal{I}_{n,k}=(-1)^k\sin(k \pi/2n)\frac{2n \cos(\pi/2n)}{n^2-1}\\ \mathcal{I}_{n,0}=\frac{-1+n \sin(\pi/2n)}{n^2-1}\\ \mathcal{I}_{n,n}=(-1)^n\mathcal{I}_{n,0} $$
Using $\Im(e^{ix})=\sin(x)$ together with the finite geometric series we easily obtain:
$$ \sum_{k=1}^{n-1} (-1)^k \mathcal{I}_{n,k}=\cot(\pi/2n)\frac{2n \cos(\pi/2n)}{n^2-1} $$
so from $(\star)$ we have
$$ \pi I_n= \frac{2}{n^2-1}\left(\frac{n}{\sin(\pi/2n)}-1\right) $$
The limit $n \rightarrow \infty$ is now easily obtained:
$$ I_{\infty}=\frac{4}{\pi^2} $$ as conjectured.
Partial results II: $m=2$, $n \in 2\mathbb{N}+1$.
Since $\sin(2x)$ and $\cos(nx)$ have both a nice symmetry point (both change sign simultaniously in the same direction) at $x=\frac{\pi}2$ we can split the integral in this case as follows:
$$ \pi I_{2,n}= 2(\mathcal{I}_{2,n,0}+\sum_{k=1}^{\lfloor{n/2}\rfloor} (-1)^k \mathcal{I}_{2,n,k})\quad(\star \star) $$
we, again by Lemma 1, have:
$$ \mathcal{I}_{2,n,k}=(-1)^k\sin(2 k \pi/n)\frac{2n \cos(\pi/n)}{n^2-4}\\ \mathcal{I}_{2,n,0}=\frac{-2+n \sin(\pi/n)}{n^2-4}\\ $$
Using once again $\Im(e^{ix})=\sin(x)$ together with the finite geometric series we easily obtain:
$$ \sum_{k=1}^{\lfloor{n/2}\rfloor} (-1)^k \mathcal{I}_{2,n,k}=\frac{\cot(\pi/2n)}{2}\frac{2n \cos(\pi/n)}{n^2-4} $$
merging everything in $(\star \star)$ we get (after some trig-manipulations)
$$ \pi I_{2,n}= \frac{2}{n^2-4}\left(n\cot(\pi/2n)-2\right) $$
Taking the limit of $n\rightarrow\infty$ we get, since $\cot(y)\sim_{0_+} 1/y$
$$ I_{2,\infty}=\frac{4}{\pi^2} $$ as expected.
More tomorrow
Assume that $m\ge 1, n\ge 1$, $m\ne n$ and $\mathrm{gcd}(m, n) = 1$.
By the identity $$\left|\sin x\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k x),$$ we have $$|\sin mt | = \frac{2}{\pi}-\sum_{i = 1}^\infty \frac{4}{\pi(4i^2-1)}\cos(2i mt)$$ and $$|\cos nt| = |\sin (\pi/2 - nt)| = \frac{2}{\pi}-\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}(-1)^j \cos(2jnt).$$ Then we have \begin{align} |\sin (mt) \cos (nt)| &= \frac{4}{\pi^2} - \frac{2}{\pi}\sum_{i = 1}^\infty \frac{4}{\pi(4i^2-1)}\cos(2i mt)\\ &\quad - \frac{2}{\pi}\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}(-1)^j \cos(2jnt)\\ &\quad + \sum_{i = 1}^\infty \sum_{j = 1}^\infty \frac{16(-1)^j}{\pi^2(4i^2-1)(4j^2-1)}\cos(2i mt)\cos(2jnt). \end{align} Note that $$\int_0^\pi \cos(2i mt)\cos(2jnt) \mathrm{d} t = \left\{\begin{array}{ll} 0 & im - jn \ne 0 \\[4pt] \frac{\pi}{2} & im - jn = 0. \end{array} \right.$$ Then we have \begin{align} \frac{1}{\pi}\int_0^\pi |\sin (mt) \cos (nt)| \mathrm{d} t &= \frac{4}{\pi^2} + \sum_{i = 1}^\infty \sum_{j = 1}^\infty \frac{8(-1)^j}{\pi^2 (4i^2-1)(4j^2-1)}\delta(i m - jn)\\ &= \frac{4}{\pi^2} + \sum_{k=1}^\infty \frac{8(-1)^{km}}{\pi^2 (4k^2n^2-1)(4k^2m^2-1)}\\ &= \frac{4}{\pi^2} + \frac{1}{m^2n^2}\sum_{k=1}^\infty \frac{8(-1)^{km}}{\pi^2 (4k^2-1/n^2)(4k^2-1/m^2)}. \end{align} Note that $$\left|\sum_{k=1}^\infty \frac{8(-1)^{km}}{\pi^2 (4k^2-1/n^2)(4k^2-1/m^2)} \right| \le \sum_{k=1}^\infty \frac{8}{\pi^2 (4k^2-1)(4k^2-1)} = \frac{\pi^2 - 8}{2\pi^2}.$$ Thus, we have $$\lim_{\max(m, n) \to \infty} \frac{1}{\pi}\int_0^\pi |\sin (mt) \cos (nt)| \mathrm{d} t = \frac{4}{\pi^2}.$$