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Q1 :
Let P' be the number trucated after 1000 digits.

Since 16\,\mid\,10^4, it is sufficient to check the last four digits of P' :
1-9     : 1*9 = 9 digits
10-99   : 90*2 =180 digits
100-370 : 271*3= 813 digits

So P' ends in \ldots 3683693 and the last four digits are 3693
3693=16\times 230+\boxed{13}

So the remainder is 13 when P' is divided by 16.

Q2 :
The remainder will be same as the remainder when sum of digits of P is divided by 9(we don't need to worry about placevalue here as 10^n\equiv 1\pmod{9} for all n\in\mathbb{Z^+}) :
1+2+3+\cdots+1000=\dfrac{1000(1000+1)}{2}=500500\equiv 5+5=10\equiv\boxed{1}

Therefore the remainder is 1 when P is divided by 9.
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