Let $a:=\alpha$ and $u:=\frac1{5\theta}$. We shall assume that $0<u<1/2$; otherwise, the problem does not seem to make much sense. We need to compute \begin{equation} \min_{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,u\le a,\\ 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) \log (1-a)-30 a \log (a)-410 u+9}{30 (a-1)}, \end{equation} \begin{equation} F_2(u,a):=\frac{H(u,a)}{60(a-1)^2 a}, \end{equation} \begin{multline} H(u,a):=-60 a^3 \log (a)-820 a^2 u-213 a^2+60 a^2 \log (a)+820 a u+78 a \\ +60 (a-1) a \log \left(\frac{1}{2} \left(\frac{1}{a}-1\right)\right)+60 (a-1)^2 a \log (1-a)+15. \end{multline} So, the minsup in question is \begin{equation} \min_{1/5<u<1/2}M_{12}(u)\wedge\min_{0<u\le1/5}(M_{11}\vee M_2(u)), \end{equation} \begin{equation} M_{11}:=\sup_{1/5<a\le1/2}F_1(u,a),\quad M_{12}(u):=\max_{a\in[u,1/2]}F_1(u,a),\quad M_2(u):=\max_{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 $DF_1(a)=-3/10 + 41 u/3 + \log a$ is increasing in $a$. So, $DF_1(a)$ (and hence $F_1'(a)$) can change the sign only from $-$ to $+$. So, \begin{align} M_{11}&=\sup_{1/5<a\le1/2}F_1(u,a)=F_1(u,1/5)\vee F_1(u,1/2),\\ M_{12}(u)&=\max_{u\le a\le1/2}F_1(u,a)=F_1(u,u)\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 $DF_2'(a) 2 (1-a)^2 a^3=1 - 5 a + 11 a^2 - 17 a^3 + 2 a^4>0$ 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)=\max_{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}
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So, the minsup in question is \begin{equation} \inf_{1/5<u<1/2}[F_1(u,u)\vee F_1(u,1/2)] \bigwedge\min_{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} We have
Lemma 1. $F_1(u,u)\le F_1(u,1/2)$ if $1/5<u<1/2$.
Lemma 2.
\begin{equation} \inf_{1/5<u<1/2}F_1(u,1/2)=F_1(1/5,1/2)=-3/5 + 82 (1/5)/3 - \log 4=3.4803\dots. \end{equation}
Lemma 3. \begin{equation} \min_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=3.7373\ldots>3.4803\dots. \end{equation}
These lemmas will be proved at the end of this answer. Since $F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)\ge F_2(u,u)$, it follows from these lemmas that
the minsup in question is $-3/5 + 82 (1/5)/3 - \log 4=3.4803\dots$, "attained in the limit" at $u=1/5$ and $a=1/2$.
It remain to prove Lemmas 1,2,3.
Proof of Lemma 1. Indeed assume that $1/5<u<1/2$. We have to check that \begin{equation} d_1(u):=F_1(u,1/2)-F_1(u,u)\ge0. \end{equation} Let $Dd_1(u):=d1'(u) 30 (1-u)^2=419 - 1640 u + 820 u^2 - 30 \log u$. Then $(Dd_1)'(u)=10(-3 - 164 u + 164 u^2)/u<0$. So, $Dd_1(u)$ is decreasing. So, $Dd_1(u)$ (and hence $d_1'(u)$) can change the sign only from $+$ to $-$. Also, $d_1(1/5)>0=d_1(1/2)$. So, $d_1(u)\ge0.\quad$ $\Box$
Proof of Lemma 2.
We have $F_1(u,1/2)=-3/5 + 82 u/3 - \log 4$, which is increasing in $u$. So,
\begin{equation}
\inf_{1/5<u<1/2}F_1(u,1/2)=-3/5 + 82 (1/5)/3 - \log 4=3.4803\dots.\quad \Box
\end{equation}
Proof of Lemma 3. For $0<u\le1/5$, consider $g(u):=F_2(u,u)$ and then $Dg(u):=g'(u)(1 - u)^2$. Then $(Dg)'(u)$ is a simple rational function of $u$, which is $\ge0$. So, $Dg(u)$ is increasing in $u$. So, $Dg(u)$ (and hence $g'(u)$) can change the sign only from $-$ to $+$. Now we find \begin{equation} \min_{0<u\le1/5}F_2(u,u)=\min_{0<u\le1/5}g(u)=3.7373\ldots>3.4803\dots. \quad \Box \end{equation}