$1.$

$aH_2SO_4+bAl_2O_3 \to cAl_2(SO_4)_3+dH_2O$

Conservation of element $H:$$2a=2d$`=>`$a=d$

Conservation of element $S:$$a=3c$

Conservation of element $Al:2b=2c$`=>`$b=c$

Conservation of element $O:$$4a+3b=12c+d$

`=>`$4d+3c=12c+d$

`=>`$d=3c$

`=>`$c=1;d=3$

`=>`$a=3;b=1$

So, $3H_2SO_4+Al_2O_3 \to Al_2(SO_4)_3+3H_2O$

$2)$

$\text{ x 1 |}$$\mathop{Cu}\limits^{0} \to \mathop{Cu}\limits^{+2}+2e$

$\text{ x 2 |}$$\mathop{Ag}\limits^{+1}+1e \to \mathop{Ag}\limits^{0}$

`=>`$Cu+2AgNO_3 \to Cu(NO_3)_2+2Ag$

$3)$

$aCH_4+bO_2 \to cCO_2+dH_2O$

Conservation of element $C:$$a=c$

Conservation of element $H:$$4a=2d$

`=>`$d=2a$

Conservation of element $O:$$2b=2c+d$

`=>`$2b=2a+2a$

`=>`$b=2a$

`=>`$a=1;b=2$

`=>`$c=1;d=2$

So, $CH_4+2O_2 \to CO_2+2H_2O$

$4)$

$aHI+bKOH \to cKI+dH_2O$

Conservation of element $I:$$a=c$

Conservation of element $O:$$b=d$

Conservation of element $H:$$a+b=2d$`=>`$2a=2d$`=>`$a=d$

Conservation of element $K:$$b=c$

`=>`$a=b=c=d=1$

So, $HI+KOH \to KI+H_2O$

Answer:

Count the atoms on each side. if some atoms are in a fixed group, count that group on each side. Generally, we balance the number of metal atoms. Oxygen is hardly ever initially treated, except some easy reactions. When we balance a reaction, we try to balance the number of every element's atom on each side by adding a number in front of all the formulas which is called coefficent.

1.

On the right side we have two aluminium atoms and also two on the left side. On the right side we have three sulfate ($SO_4$) groups and one sulfate group on the left side, so we put a $1$ in front of $Al_2(SO_4)_3$ and a $3$ in front of $H_2SO_4$. Put a $1$ in front of $Al_2O_3$ to balance the number of aluminum atoms. Put a $3$ in front of $H_2O$ to balance the number of hydrogen atoms. To have this checked, count the number of oxygen atoms on each side.

Balanced reaction: $3H_2SO_4+Al_2O_3\to Al_2(SO_4)_3+3H_2O$. Coefficent $1$ isn't often written.

2.

Follow those steps, the result is $Cu+2AgNO_3 \to Cu(NO_3)_2+2Ag$. At first, we notice that nitrate group is not balanced yet, so we balance the number of nitrate groups, silver atoms and copper atoms.

3.

The number of carbon atoms is already equal on both sides. Balance the number of hydrogen atoms and then oxygen to complete the reaction.

$CH_4+2O_2\to CO_2+2H_2O$

4.

The number of potassium atoms is already equal on both sides, the same to iodine. Count the number of hydrogen atoms and oxygen atoms, they are also equal on both sides. This means that all the coefficent of this reaction is $1$, we don't need to put any numbers.

$HI+KOH\to KI+H_2O$