Let $a:=\alpha$ and $u:=\frac1{5\theta}$. The condition $\theta\ge1$ (now added in the question) means that $0<u\le1/5$, which will be assumed henceforth. 
We need to compute 
\begin{equation}
	\inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), 
\end{equation}
where
\begin{equation*}
	F(u,a):=\begin{cases}
	F_1(u,a)&\text{ if }1/5<a\le1/2,\\
	F_2(u,a)&\text{ if }u\le a\le1/5,
	\end{cases}  
\end{equation*}
\begin{equation}
	F_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2},
\end{equation}
\begin{equation}
	F_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2},  
\end{equation}
\begin{multline}
	H(u,a):=-60 a^3 \ln a-820 a^2 u \ln2+60 a^2 \ln a-213 a^2 \ln2+820 a u \ln2 \\ 
	+60
   (a-1) a \ln \left(\frac{1}{2} \left(\frac{1}{a}-1\right)\right)+60 (a-1)^2 a \ln
   (1-a)+78 a \ln2+15 \ln2. 
\end{multline}
So, the infsup in question is
\begin{equation}
	\inf_{0<u\le1/5}(M_1(u)\vee M_2(u)), 
\end{equation}
\begin{equation}
	M_1(u):=\sup_{1/5<a\le1/2}F_1(u,a),\quad
	M_2(u):=\sup_{a\in[u,1/5]}F_2(u,a). 
\end{equation}	
 
For $F_1(a):=F_1(u,a)$, let $DF_1(a):=F_1'(a)(1-a)^2$. Then $DF_1(a)=-3/10 + 41 u/3 + \ln a/\ln2$ is increasing in $a$. So, $DF_1(a)$ (and hence $F_1'(a)$) can change the sign only from $-$ to $+$. So, 
\begin{align}
	M_1(u)&=\sup_{1/5<a\le1/2}F_1(u,a)=F_1(u,1/5)\vee F_1(u,1/2). 
\end{align}

For $F_2(a):=F_2(u,a)$, let $DF_2(a):=F_2'(a)(1-a)^2$. Then $DF_2'(a) 2 \ln2\,(1-a)^2 a^3=2 a^4 + \ln2 - a (2 + \ln8) + a^2 (8 + \ln8) - 
 a^3 (8 + \ln512)>0$ for $a\in[0,1/5]$. So, $DF_2(a)$ is increasing in $a\in[0,1/5]$, and so, $DF_2(a)$ (and hence $F_2'(a)$) can change the sign only from $-$ to $+$. So, 
\begin{equation}
	M_2(u)=\sup_{a\in[u,1/5]}F_2(u,a)=
	F_2(u,u)\vee F_2(u,1/5)\quad\text{if }0<u\le1/5. 
\end{equation}



So, the infsup in question is
\begin{equation}
\inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)].  
\end{equation}
Since $F_1(u,1/5), F_1(u,1/2), F_2(u,1/5)$ are affine in $u$, it is easy to see that $F_1(u,1/5)\vee F_1(u,1/2)< F_2(u,1/5)$ if $0<u\le1/5$. So,  
the infsup in question is
\begin{equation}
\inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)].  
\end{equation}
For brevity, let 
\begin{equation}
	g(u):=F_2(u,u),\quad d(u):=F_2(u,u)-F_2(u,1/5)=g(u)-F_2(u,1/5). 
\end{equation}
Let $g_1(u):=g'(u)(1 - u)^2$. Then $g_1'(u)$ is a simple rational function of $u$, which is $>0$ (everywhere here $0<u\le1/5$). So, $g_1(u)$ is increasing in $u$. So, $g_1(u)$ (and hence $g'(u)$) can change the sign only from $-$ to $+$. Now we find 
\begin{equation}
	\inf_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=2.8343\ldots, 
\end{equation}
attained at $u=0.095260\ldots$. 

Because (i) $d(u)=g(u)-F_2(u,1/5)$, (ii) $F_2(u,1/5)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d'(u)$ can change the sign only from $-$ to $+$. Also, $d'(13/100)=-6.4029\ldots<0$. So, $d$ is decreasing on $[0,13/100]$, with $d(13/100)=0.10364\ldots>0$. So, $d>0$ on $[0,13/100]$, that is, $F_2(u,u)>F_2(u,1/5)$ if $0<u\le13/100$. Thus, 
\begin{equation}
\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= \min_{0<u\le13/100}F_2(u,u)= 2.8343\ldots,  
\end{equation}
attained at $u=0.095260\ldots$. 
On the other hand, because $F_2(u,1/5)]$ is increasing in $u$, we have 
\begin{multline}
\inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\ge\inf_{13/100<u\le1/5}F_2(u,1/5) \\ 
= 
F_2(13/100,1/5) =2.9434\ldots>2.8343\ldots
=\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]. 
\end{multline}

> Thus, the infsup in question is
\begin{multline}
\inf_{0<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)] \\ 
=\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]\bigwedge
\inf_{13/100<u\le1/5}[F_2(u,u)\vee F_2(u,1/5)]\\
=\inf_{0<u\le13/100}[F_2(u,u)\vee F_2(u,1/5)]= 2.8343\ldots, 
\end{multline}
attained at $a=u=0.095260\ldots$.