Hint $ $ If $ \,x,y\,$ are solutions then $ \ cx\equiv b\equiv cy\pmod{\! m}\ $ so, defining $ \ d=(m,c)\,$ we have
$$ \,m\,|\,c\,(x\!-\!y) \iff \color{#0a0}{\frac{m}d}\,\bigg|\,\color{#0a0}{\frac{c}d}\,(x\!-\!y)\color{#C00}{\iff} \frac{m}d\,\bigg|\ x\!-y,\ \ {\rm by} \ \ \left(\color{#0a0}{\frac{m}d,\frac{c}d}\right)\color{#90f}{= \frac{(m,c)}d} = 1\qquad$$
Remark $\ $ The final $ $ '$\color{#C00}{\!\!\iff}\!\!$' $ $ employs $\rm\color{#0a0}{Euclid's\ Lemma}$ and the $\rm\color{#90f}{\rm distributive\ law}$ for GCDs.
Alternatively $\ $ Bezout yields: $ \, d = (c,m) = j\:\!c+k\:\!m,\,\ j,k\in\Bbb Z.\,$ Let $ \,z = x-y.\,$ Then
$\!\bmod m\!:\ \color{#0a0}{cz\equiv 0\equiv mz}\ \Rightarrow\ dz = (c,m)z = j\:\!\color{#0a0}{cz}+k\:\!\color{#0a0}{mz \equiv 0},\,$ so $ \ m\,|\,dz\,\Rightarrow\,m/d\,|\,z$
Intuitively see this answer for an approach using modular fractions.
Using this, $(10,25)=5$ must divide $9$ to admit any solution, which is not, so there is no solution.
$3x\equiv 24\pmod 6\implies 2\mid x$ i.e., any even value(not only $8$) of $x$ will satisfy the 1st congruence.
So, the system of the linear congruence has no solution as the 2nd congruence is not solvable.
Alternatively, for the 2nd congruence, $10x-18=25y$ for some integer $y$,
So,$5(2x-5y)=18, \frac{18}5=2x-5y$ which is an integer, hence contradiction.
Best Answer
It seems like you're familiar with the theorem in the comments above. For part (a), by inspection, you can see that $x\equiv 2$ is a solution. Since $g=\gcd(10,15)=5$ and $m/g=15/5=3$, you know there are $5$ total solutions $\pmod{15}$, and the others are found just be adding $3$ successively until you've found all $5$.
For (b), $\gcd(6,26)=2$ but $2\nmid 7$, so how many solutions can there be?
Part (c) is nice because $7$ and $11$ are coprime, so $7$ is actually invertible here. Try to find $7^{-1}\pmod{11}$, and then multiply both sides of $7x\equiv 8\pmod{11}$ by it to find the unique solution for $x$ modulo $11$.