For the first one, it helps to know that if $a,b$ are relatively prime then
any prime factor of $a^4+b^4$ is either $2$ or congruent to $1 \bmod 8$.
(This happens to be in my first published paper $-$ admittedly this
"publication" was the proceedings of a local high-school Math Fair...
$-$ and can proved similarly to one proof of the corresponding $1 \bmod 4$
result for $a^2+b^2$, by finding an $8$th root of unity $a/b \bmod p$.)
Since $n$ and $n+1$ are relatively prime and of opposite parity,
this means you need only check divisibility by $17$, $41$, $73$, etc.
For $n < 5$, the value of $n^4 + (n+1)^4$ is small enough that
the only candidate composite number is $17^2 = 289$, which you can
exclude by direct computation or otherwise (e.g. Fermat proved
that there's no solution of $a^4+b^4=c^2$ in positive integers;
or, the only Pythagorean triangle with hypotenuse $17$ has sides
$8$ and $15$). So $n=5$ is the first candidate, and the first
candidate prime factor actually divides $5^4 + 6^4 = 1921 = 17 \cdot 113$,
so we're done.