Column C

First off you'll notice the text "Column C to be used in all cases except as otherwise permitted in Note 3.)", in the title of table 220.55. This makes it simple. You have 2 appliances, so follow that over in the table, and you'll see 11 in Column C. So there you go, you can just use 11 kW. Done.

11,000 W / 240 V = 45.8333 A

So you'll need a 50 ampere breaker, and wire sized appropriately for the load.

Note 3

Note 3 says:

3. Over 1 3⁄4 kW through 8 3⁄4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances. Where the rating of cooking appliances falls under both Column A and Column B, the demand factors for each column shall be applied to the appliances for that column, and the results added together.

Perfect, so instead of just using the value from column C you can do math. Let's step through it.

...it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW...

8.4 kW + 5.0 kW = 13.4 kW

...and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances...

Let's check the table again... You have 2 appliances, both between 3 1/2 and 8 3/4 kW. So You'll look at column B, and find 65%.

13.4 kW * 65% = 8.71 kW

8710 W / 240 V = 36.2916 A

So using this method you can use a 40 ampere breaker, and appropriately sized wire. However, keep in mind that if you change the equipment, you'll have to do the calculation again. So while you can use this value, you may have to upgrade the circuit later if you change equipment.

Note 4

I'm not exactly sure how note 4 comes into play, but I think it can be used if this is the only equipment on the branch circuit. Just for fun, let's run through that one too.

4. Branch-Circuit Load. It shall be permissible to calculate the branch-circuit load for one range in accordance with Table 220.55. The branch-circuit load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branch-circuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be calculated by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range.

You're only concerned with the second half of this note, since you have one counter-mounted cooking unit, and one wall-mounted oven, all supplied by a single branch-circuit, and located in the same room. So you can add the nameplate values, and treat it as a single range.

8.4 kW + 5.0 kW = 13.4 kW

So you can treat the units as a single 13.4 kW range. Check the column C again, this time for a single range. You'll find a value of 8 kW. But wait... The column header says "(Not over 12 kW Rating)". Your range is 13.4 kW. That's bigger than 12 kW. Now you'll have to check note 1

1. Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.

That's easy enough.

13.4 kW - 8 kW = 5.4 kW

Since .4 is not a "major fraction", you can just use 5 kW. So you'll have to add 5% 5 times.

5% * 5 = 0.25
8000 W * 0.25 = 2000 W
8000 W + 2000 W = 10,000 W

That means you'll have to use 10 kW as your demand.

10,000 W / 240 V = 41.666 A

Which means you can use a 50 ampere breaker, and appropriately size wire.

As one of the commenters pointed out, this is not 2-phase power (at least assuming something similar to standard U.S. residential 240V single-phase/split-phase service--what country are you in?). It's single-phase power. You get 240V between the poles on the smaller coil of the transformer outside your place. So you get 120V (half) from each pole to the exact center of the coil, where the "neutral" taps in (it's a single-phase center tap transformer). The neutral carries the imbalance between the loads on the two poles (power from one pole is opposite polarity and thus cancels out power from the other). So if there's exactly the same load on each pole, the neutral carries zero current, but it almost always carries some current.

So... yes, technically this would work. But where are you going to tap in? With what kind of device? Are you planning to connect a #8 and #12 wire with a wire nut? ;-)

As another commenter pointed out, you will end up with a 120V appliance running on a circuit protected by a 40A breaker. Since you can't legally put a 15A or 20A outlet on a 40A circuit, that definitely violates code. The device itself isn't going to draw any more current than it needs unless it has a short, but you can't legally wire in an outlet that the appliance cord will fit.

Your 37A calculation is 93% of the rating of the 40A circuit. U.S. NEC says you can't design a circuit with a continuous load over 80% of the circuit breaker's capacity. But "continuous" means full draw for longer periods of time than you are going to operate your stove.

And your cooler only draws about one Amp. So you might be okay there.

So... if you absolutely, positively can't or won't pull a new 120V 15A or 20A circuit from the panel, it sounds like the sanest option (from yet another commenter) is to redirect that 40A 240V circuit into a small subpanel.

I just now saw a little 2 space, 4 circuit 70A capacity subpanel/load center at a major box store's website for $15. Cheap.

Hopefully the cable you already have running to the stove has 4 wires (2 hot, 1 neutral, 1 ground).

The neutral (grounded) and grounding buses in the subpanel have to be disconnected from each other, the neutral bus has to be isolated from the metal box, and the grounding bus has to be bonded to the box. Otherwise, if there's a fault in the neutral, all your return current ends up on the ground and potentially energizes the box.

Your existing 40A two-pole breaker in the main panel is your master breaker for the subpanel, protecting the existing wire. Put a 40A two-pole breaker in the subpanel to feed the stove, and a 15A or 20A single-pole breaker to feed an outlet for the cooler.

You can use a masonry bit and concrete screws to drill and screw the little subpanel to the concrete, and to secure the conduit or MC that you'll run from that panel to the outlets for the stove and cooler. Which you will secure to the concrete in similar fashion.

You could use a small panel similar to this: