In this case the numbers are small enough that the easiest way to find $a_1$ and $b_1$ is trial and error: I notice immediately that $4\cdot5$ and $3\cdot7$ differ by $1$, which gives my an $a_1$ and $b_1$ that work.
In general if $\gcd(m,n)=d$, there are integers $a$ and $b$ such that $am+bn=d$; in this problem $d=1$. If you apply the Euclidean algorithm to $m$ and $n$ to get the gcd $d$, you can then back-solve to write $d$ in the form $am+bn$; this gives you a systematic approach to finding $a_1$ and $b_1$. Here, for instance, the Euclidean algorithm gives you
and I can take $a_1=3,b_1=-2$, an even nicer choice than my top-of-the-head choice in the first paragraph.
The second question is easily answered by induction. If you can find solutions for $5$ consecutive numbers (e.g., $24,25,26,27$, and $28$, then there must be a solution for each integer larger than these, since each larger integer is one of these plus a multiple of $5$. This provides an alternative proof for the first problem, by the way, since you can also repeatedly subtract $5$ from these five solutions to get solutions, possibly in negative integers, for every smaller integer as well, including the negative integers. This requires another induction, this time a downward induction through the integers less than the starting five.
The general result is that for relatively prime integers $m$ and $n$, every integer greater than or equal to $(m-1)(n-1)$ can be represented as $am+bn$ for non-negative integers $a$ and $b$. (In your case $(5-1)(7-1)=24$.) Moreover, exactly half of the non-negative integers less than $(m-1)(n-1)$ can be so represented, and $(m-1)(n-1)-1$ is one that cannot. Links in this answer will give you more information on this.
Added to answer a question in the comments: The Euclidean algorithm applied to $183$ and $257$ yields
Back-solving for $1$ in terms of $183$ and $257$:
And indeed $47\cdot257-66\cdot183=12079-12078=1$.