I1.1 The obtuse angle formed by the hands of a clock at $10: 30$ is $(100+a)^{\circ}$. Find $a$.
11.2 The lines $a x+b y=0$ and $x-5 y+1=0$ are perpendicular to each other. Find $b$.

I1.3 If $(b+1)^{4}=2^{c+2}$, find $c$.
I1.4 If $c-9=\log _{c}(6 d-2)$, find $d$.
At 10:00, the angle between the arms of the clock $=60^{\circ}$
From 10:00 to 10:30, the hour-hand had moved $360^{\circ} \times \frac{1}{12} \times \frac{1}{2}=15^{\circ}$.
The minute hand had moved $180^{\circ}$.
$$
\begin{array}{l}
100+a=180-60+15=135 \Rightarrow a=35 \\
-\frac{35}{b} \times \frac{1}{5}=-1 \\
\Rightarrow b=7 \\
8^{4}=2^{c+2} \\
2^{3(4)}=2^{c+2} \\
\Rightarrow c=10 \\
10-9=1=\log _{10}(6 d-2) \\
\Rightarrow 6 d-2=10 \\
\Rightarrow d=2
\end{array}
$$