• For (a), use $a\mid b$ and $a\mid c$ implies $a\mid b-c$.
• For (b), deduct that $73\mid n$ or $13\mid n-1$.  AND $9\mid n$ or $27\mid n-1$. This gives 4 cases.

a) 

No. For the sake of contradiction, assume that $3\mid n$ and $3\mid n-1$

So, $3\mid n-(n-1)\implies 3\mid 1$. Contradiction.

b) 

The answer is $n=585$.

$1971=3^3\cdot 73$. So, $73\mid n^2\implies73\mid n$ or $73\mid n-1$.

$27\mid n^2\implies 9\mid n$ or $9\mid n-1$.

Case 1: 

$73\mid n-1$ and $9\mid n$.

Let $n-1=k\cdot 73$.

$73\cdot k\equiv n-1 \equiv 8\pmod{9}$.

But, $73\equiv 1\pmod{9}$

So, $k\equiv 8\pmod9$. Let $k=9l+8$

But, for $l=0$, $n=585$ works.

Case 2: 

$73\mid n$ and $27\mid n-1\implies 9\mid n-1$.

Let $n=k\cdot 73$.

$73\cdot k\equiv n\equiv (n-1)+1 \equiv 1\pmod{9}$.

But, $73\equiv 1\pmod{9}$

So, $k\equiv 1\pmod9$. Let $k=9l+1$

But, for $l=0$, $n=73$ does not work. For $l>0$, $n>10\cdot 73 = 730>585$.

Case 3: $73\mid n$ and $9\mid n$.

So, $9\cdot 73\mid n\implies n\geq 657>585$

Case 4: $73\mid n-1$ and $27\mid n-1$.

So, $27\cdot 73\mid n-1\implies n>n-1\geq 1971>585$

So, the answer 585 is proved.