The problem states that the cost C is partly constant and partly varies as the square root of the number n of students. This can be written as:

The problem states that the cost $C$ is partly constant and partly varies as the square root of the number $n$ of students. This can be written as:

$$C = k_1 + k_2 \sqrt{n}$$

where $k_1$ is the constant part and $k_2$ is the constant of proportionality.

We are given two data points:

1. When $n = 144$, $C = 640$.
2. When $n = 64$, $C = 560$.

Substitute these values into the equation:
For $n = 144$:

$$640 = k_1 + k_2 \sqrt{144}$$ $$640 = k_1 + 12k_2 \quad (1)$$

For $n = 64$:

$$560 = k_1 + k_2 \sqrt{64}$$ $$560 = k_1 + 8k_2 \quad (2)$$

Subtract equation (2) from equation (1):

$$(640 - 560) = (k_1 + 12k_2) - (k_1 + 8k_2)$$ $$80 = 4k_2$$ $$k_2 = \frac{80}{4}$$ $$k_2 = 20$$

Substitute $k_2 = 20$ into equation (2):

$$560 = k_1 + 8(20)$$ $$560 = k_1 + 160$$ $$k_1 = 560 - 160$$ $$k_1 = 400$$

a) Equation connecting C and n:
Substitute the values of $k_1$ and $k_2$ back into the general equation:

$$C = 400 + 20\sqrt{n}$$

The equation connecting $C$ and $n$ is $\boxed{C = 400 + 20\sqrt{n}}$.

b) Cost of feeding 196 students:
Substitute $n = 196$ into the equation found in part (a):

$$C = 400 + 20\sqrt{196}$$ $$C = 400 + 20(14)$$ $$C = 400 + 280$$ $$C = 680$$

The cost of feeding 196 students is \boxed{\680.00}$.

**c) Number of students that can be fed with $700.00:** Substitute$C = 700$ into the equation found in part (a):

$$700 = 400 + 20\sqrt{n}$$ $$700 - 400 = 20\sqrt{n}$$ $$300 = 20\sqrt{n}$$ $$\sqrt{n} = \frac{300}{20}$$ $$\sqrt{n} = 15$$

Square both sides to find $n$:

$$n = 15^2$$ $$n = 225$$

The number of students that can be fed with $700.00 is \boxed{225 \text{ students}}$.