Normal approximation:
Consider a single card draw from the deck:
E(X)=7
Var(X)=14
With the sum of 5 cards (drawn without replacement):
$E(X)=7 \times 5$
$\text{Var}(X)=14\times 5\times \frac{52-5}{52-1}$
(the last term being the 'finite population correction', which applies as much to the variance of the sum as it does to the variance of the mean).
The approach to normality in this situation is reasonably rapid.
This suggests the sum on 5 cards might be be roughly approximated by a normal distribution with mean $35$ and variance $70\times\frac{47}{51}\approx 64.5098$. Using a continuity correction this gives an approximate probability of totalling at least 40 on 5 cards of:
$$1-\Phi(\frac{39.5-35}{\sqrt{70\times\frac{_{47}}{^{51}}}})\approx 0.288$$
Simulation (in R) of ten million five-card draws indicates the probability is
about $0.293$ (with s.e. $\approx 1.4\times 10^{-4}$):
res=replicate(10000000,sum(sample(rep(1:13,4),5,replace=FALSE)));mean(res>=40)
[1] 0.2927447
As a check on the earlier calculation of the variance, the standard deviation of of those simulated sums was 8.0316; the previous calculation gives 8.0318.
Edit: Here's a comparison of the empirical cdf of the simulated data with the above normal approximation; they're pretty close:
More extensive simulations are consistent with the other two answers based on complete enumeration:
Just sum the probabilities of $1-2, 2-3, 3-4,..., 12-13$
probability to get $1-2$ = $2* \frac{4}{52}*\frac{4}{51}$
The total probability = $\frac{2*12*4*4}{52*51}$
Best Answer
Your question about drawing a $4,6$ or a $6,4$ is a good one.
I'd break it down into two cases: one with the cards the same, the other with cards different.
If the cards are the same, first choose the face value of the card ($13$) then choose the two suits ($_4C_2 = 6$). So there are $13 \cdot 6 = 78$ possible ways to draw a pair. Of these, two $5$'s through two $8$'s satisfy your criterion: $4$ values $\times 6$ suit combinations for each $= 24$ possibilities.
If the cards are different, first choose the two face values ($_{13}C_2 = 78$) then choose the suit for each card ($16$). So there are $78 \cdot 16 = 1248$ ways to draw two different cards.
(Note: $_{52}C_2 = 1326 = 78 + 1248,$ so we've indeed accounted for all possibilities.)
Counting the number of possibilities with the sum between $10$ and $16$ with two different cards is bookkeeping. I'll start with the larger card, and work my way up from the bottom with the smaller cards to go through the possibilities methodically. I'll also not write a higher card second, because that would double-count.
$A2, A3, A4, A5$
$K2, K3, K4, K5, K6$
$Q2, Q3, Q4, Q5, Q6$
$J2, J3, J4, J5, J6$
$T2, T3, T4, T5, T6$
$92, 93, 94, 95, 96, 97$
$82, 83, 84, 85, 86, 87$
$73, 74, 75, 76$
$64, 65.$
I count $42$ combinations for the values. Since either card can be any suit, that's $42\cdot 16 = 672$ possible combinations.
Now we're basically done:
$$P = \frac{672+24}{1326} \approx 0.525.$$