Number of solutions of $x_1 + x_2 + x_3 + x_4 = 14$ such that $x_i \le 6$
The answer you got for the first question is right.
For the second, call a distribution bad if one or more of the $x_i$ is $\ge 7$. Our strategy is to count the number of bads, and subtract from the answer of a).
One can have $2$ of the $x_i$ equal to $7$. This can be done in $\binom{4}{2}$ ways.
Now we count the number of bads in which only one of the $x_i$ is $\ge 7$. Which one it is can be chosen in $\binom{4}{1}$ ways. Suppose that $x_1\ge 7$. Give $7$ candies to kid 1. The remaining $7$ have to be split between the four people, with none of 2, 3, or 4 getting $7$. There are $\binom{10}{3}-3$ ways to do this.
We get a total of $\binom{4}{1}\left[\binom{10}{3}-3\right]+\binom{4}{2}$ bads.
Alternately, we use Inclusion/Exclusion more explicitly. Choose one of the $4$ to give at least $7$ to, and give her $7$. We can distribute the remaining $7$ among the $4$ people in $\binom{10}{3}$ ways. But this double-counts the $\binom{4}{2}$ ways to give $7$ to two of the variables. So the number of bads is $\binom{4}{1}\binom{10}{3}-\binom{4}{2}$.
For part "a" it is correct.
For part "b":
Using Multinomial:
The ways are equivalent to: $$\text{ Coefficient of $x^{14}$ in }(x^0+x^1+x^2+...x^6)^4\\ =\text{ Coefficient of $x^{14}$ in }(1-x^7)^4(1-x)^{-4}\\ =\text{ Coefficient of $x^{14}$ in }\left(1-\binom41x^7+\binom42x^{14}-...\right)(1-x)^{-4}\\ =^*1\times\binom{4+14-1}{14}-\binom41\binom{4+7-1}{7}+\binom42\times1=206$$
Using the Inclusion–Exclusion Principle:
Let $p_i$ be the case such that p $x_i's$ are $\ge7$ Now: $$p_u=(p_1\cup p_2\cup p_3\cup p_4)=S_1-S_2+S_3-S_4$$ $$\begin{array}{|c|c|}\hline S_i&\text{values}\\\hline S_1&\binom41\left(\binom92+\binom82+\binom72+...\binom22\right)^{**}=480\\ S_2&\binom42\times1\times1=6\\ S_3&0\\ S_4&0\\\hline \end{array}$$ So,$p_u=480-6=474$, so that we get $680-474=206$ for b.
$^*$ Coefficient of $x^r$ in $(1-x)^{-n}$ is $\binom{n+r-1}r$
$^{**}$ Let $x_1$ be $k$ then $x_2+x_3+x_4=14-k$,ways for this are $\binom{16-k}2$
Total: $$\sum_{k=7}^{14}\binom{16-k}{2}=120$$