Let the base and altitude of the right-angled triangle be $x$ and $y \mathrm{~cm}$, respectively Therefore, the hypotenuse will be $(x+2) \mathrm{cm}$.

$\therefore(x+2)^{2}=y^{2}+x^{2}$

Again, the hypotenuse exceeds twice the length of the altitude by $1 \mathrm{~cm}$.

$\begin{array}{l}
\therefore h=(2 y+1) \\
\Rightarrow x+2=2 y+1 \\
\Rightarrow x=2 y-1
\end{array}$

Putting the value of $x$ in (i), we get:
$\begin{array}{l}
(2 y-1+2)^{2}=y^{2}+(2 y-1)^{2} \\
\Rightarrow(2 y+1)^{2}=y^{2}+4 y^{2}-4 y+1 \\
\Rightarrow 4 y^{2}+4 y+1=5 y^{2}-4 y+1 \\
\Rightarrow-y^{2}+8 y=0 \\
\Rightarrow y^{2}-8 y=0 \\
\Rightarrow y(y-8)=0 \\
\Rightarrow y=8 \mathrm{~cm} \\
\therefore x=16-1=15 \mathrm{~cm} \\
\therefore h=16+1=17 \mathrm{~cm}
\end{array}$

Thus, the base, altitude and hypotenuse of the triangle are $15 \mathrm{~cm}, 8 \mathrm{~cm}$ and $17 \mathrm{~cm}$, respectively.