Drawing without replacement - book seems wrong 
This is the book's answer.  Seems that it MUST be incorrect. This is WITH replacement.
I calculate that the prob of drawing and ace first is 1/13 and a 10 second is also 1/13.  So, it would seem that the prob of drawing an Ace first or a 10 second would be 1/13 + 1/13 - 1/169, correct?
Their math doesn't even seem to make sense with the subtrace the Pace 1st and ten 2nd) they use MY numbers not theirs!
Is the book incorrect?
 A: There are $52\cdot 51$ ways to draw the first two cards (without replacement).
Of these, $4\cdot 51$ have an Ace first (four possible aces, $51$ choices for the other card.
Of the hands without an Ace first, there are $44\cdot 4$ ways to choose a non-ten first and a ten second, and another $12$ ways to choose a ten first and a ten second.
The total is $204+176+12 = 392$ and the probability you want is 
$$
\frac{292}{2652} = \frac{98}{663}
$$
which indeed is a tad less than $\frac{2}{13}$.
EDIT Corrections have been made, pointed out by Fabio Somenzi.
A: We have several answers. This drives me to try a simulation in R statistical
software, which should be accurate to about 3 places. 
Denote 'ace on the first'
as $A$ (f.a in the simulation program), and 'ten on second' as $B$ (s.10 in program).  Then $P(A) = P(B) = \frac{204}{2652} = 0.076923$ and $P(AB) = \frac{16}{2652}.$
Hence $$P(A \cup B) = 392/2652 =  0.147813.$$
m = 10^6;  f.a = s.10 = logical(m)
deck = rep(1:13, each=4)
for (i in 1:m) {
  draw = sample(deck,2)
  f.a[i] = (draw[1]==1)    # TRUE if first is Ace
  s.10[i] = (draw[2]==10)  # TRUE if 2nd is Ten
  }
mean(f.a);  mean(s.10);  mean(f.a|s.10)
## 0.076932    # aprx P(A)
## 0.076945    # aprx P(B)
## 0.147853    # aprx P(A or B)

