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\le 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-117 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)-18 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\le 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\le 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}

Therefore and because $F_1(u,1/5)=F_2(u,1/5)$, 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)].
\end{equation}

For brevity, let
\begin{equation}
g(u):=F_2(u,u),\quad d_a(u):=F_2(u,u)-F_1(u,a)=g(u)-F_1(u,a).
\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)=1.0616\ldots,
\end{equation}
attained at $u=0.099677\ldots$.

Because (i) $d_a(u)=g(u)-F_1(u,a)$, (ii) $F_1(u,a)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d_a'(u)$ can change the sign only from $-$ to $+$. Also, $d_{1/5}'(14/100)=-6.7650\ldots<0$. So, $d_{1/5}$ is decreasing on $[0,14/100]$, with $d_{1/5}(14/100)=0.1884\ldots>0$. So, $d_{1/5}>0$ on $[0,14/100]$, that is, $F_2(u,u)>F_1(u,1/5)$ if $0<u\le14/100$. Similarly, using that $d_{1/2}'(14/100)=-17.015\ldots<0$ and $d_{1/2}(14/100)=0.07602\ldots>0$, we verify that $F_2(u,u)>F_1(u,1/2)$ if $0<u\le14/100$.

So,
\begin{equation}
\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]= \min_{0<u\le14/100}F_2(u,u)= 1.0616\ldots,
\end{equation}
attained at $u=0.099677\ldots$.
On the other hand, because $F_1(u,1/2)$ is increasing in $u$, we have
\begin{multline}
\inf_{14/100<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)]\ge\inf_{14/100<u\le1/5}F_1(u,1/2) \\
=
F_1(14/100,1/2) =1.2266\ldots \\
>1.0616\ldots
=\inf_{0<u\le14/100}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)].
\end{multline}

Thus, with $M(u):=F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)$, the infsup in question is
\begin{multline}
\inf_{0<u\le1/5}M(u)
=\inf_{0<u\le14/100}M(u)
\bigwedge
\inf_{14/100<u\le1/5}M(u)\\
=\inf_{0<u\le14/100}M(u)= 1.0616\ldots,
\end{multline}
attained at $a=u=0.099677\ldots$.