Solution:

a. Let us write $n=10 a+b$ with $a$ and $b$ decimal digits, that is $1 \leq a \leq 9$ and $1 \leq b \leq 9$: by hypothesis $a=0$ is not possible (since $n>10$), and $b=0$ is not possible because in that case, erasing the first digit of $n$ would give 0, which does not divide $n$. The conditions are then that $a$ divides $10 a+b$, which is equivalent to the fact that $a$ divides $10 a+b-10 a=b$, and that $b$ divides $10 a+b$, equivalent to $b$ dividing $10 a+b-b=10 a$. Let us then set $b=k a$, where $k$ is an integer such that $1 \leq k \leq 9$. We find that $b=k a$ divides $10 a$, that is, $k$ divides 10: if $k=1$ we find nine solutions in which $a=b$, namely $n=11,22,33,44,55,66,77,88,99$. If $k=2$ then $b<10$ implies $a<5$, and we find the solutions $n=12,24,36,48$. Finally, if $k=5$, we similarly find the unique solution $n=15$.

b. Let us write $n=10 a+b$ with $1 \leq b \leq 9$ the last digit of $n$ and $1 \leq a=(n-b) / 10$ an integer (this time not necessarily of a single digit). Erasing the last digit, we find that $a$ must divide $10 a+b$, and hence also that $a$ divides $(10 a+b)-10 \cdot a=b$. Since $b$ is a positive number less than or equal to 9, then $a$ (which divides $b$) also cannot exceed 9, so $a$ is composed of a single digit and $n$ is written with two decimal digits as desired. The solutions are then only those found in the previous point, which are all valid because they have no zero digits.